CHAPTER 9 CALCULATIONS FROM CHEMICAL EQUATIONS SOLUTIONS TO REVIEW QUESTIONS. = 2 mol PH 3. a g b = 0.4 g PH. = 12 mol H 2 O.
|
|
- Rodger Nelson
- 7 years ago
- Views:
Transcription
1 HEINS v4.qxd 1/30/06 1:58 PM Page 95 CHAPTER 9 CALCULATIONS FROM CHEMICAL EQUATIONS SOLUTIONS TO REVIEW QUESTIONS 1. A e ratio is the ratio between the e amounts of two atoms and/or ecules involved in a chemical reaction.. In order to convert grams to es the ar mass of the compound under consideration needs to be determined. 3. The balanced equation is Ca 3 P + 6 H O 3 Ca1OH + PH 3 Correct: 11 Ca 3 P PH 3 1 Ca 3 P = PH 3 Incorrect: 1 g Ca 3 P would produce 0.4 g PH 3 (d) (1 g Ca 3 P )a g b PH 3 a g b = 0.4 g PH 1 Ca 3 P 3 Correct: see equation Correct: see equation (e) Incorrect: Ca 3 P requires 1 H O to produce 4.0 PH 3. 1 Ca 3 P 6 H O 1 Ca 3 P = 1 H O (f) Correct: Ca 3 P will react with 1 H O ( 3 H O are present in excess) and 6 Ca(OH) will be formed. 1 Ca 3 P 3 Ca(OH) 1 Ca 3 P = 6 Ca(OH) (g) (h) Incorrect: (00. g Ca 3 P )a g b 6 H O a 18.0 g b = 119 g H 1 Ca 3 P O The amount of water present (100. g) is less than needed to react with 00. g Ca 3 P. H O is the limiting reactant. Incorrect: water is the limiting reactant g H Oa g b PH 3 6 H O a g b = 6.9 g PH
2 HEINS v4.qxd 1/30/06 1:58 PM Page The balanced equation is CH O + NH 3 HCN + 6 H O Correct HCN Incorrect: 116 O = 10.7 HCN (not 1 HCN) 3 O Correct (d) Incorrect: 11 HCN 6 H O (not 4 H O) HCN = 36 H O (e) Correct O (f) Incorrect: is the limiting reactant HCN 13 O = HCN 3 O (not 3 HCN) 5. The theoretical yield of a chemical reaction is the maximum amount of product that can be produced based on a balanced equation. The actual yield of a reaction is the actual amount of product obtained. 6. You can calculate the percent yield of a chemical reaction by dividing the actual yield by the theoretical yield and multiplying by one hundred
3 HEINS v4.qxd 1/30/06 1:58 PM Page 97 CHAPTER 9 SOLUTIONS TO EXERCISES 1. (d). (d) 3. (d) (e) 15.0 g KNO 3 a g b = 0.47 KNO m NaOHa b = NaOH 1000 m 15.4 * 10 g (NH 4 ) C O 4 a g b = 4.4 (NH 4) C O 4 The conversion is: ml sol g sol g H SO 4 H SO ml solutiona 1.77 g ml b g H SO 4 a 1 g solution g b = 0.37 H SO kg NaHCO 3 a 1000 g ba 1 kg g b = 5.0 NaHCO 3 1 g mg ZnCl a ba 1000 mg g b = 3.85 * 10-3 ZnCl 19.8 * ecules CO 6.0 * 10 3 ecules = 16 CO 150 ml C H 5 OHa g ml 1.55 Fe(OH) 3 a g b = 73 g Fe(OH) kg CaCO 3 a 1000 g b = 1.5 * 10 5 g CaCO kg NH 3 a g b = 179 g NH 3 1 ba g b = 4.3 C H 5 OH 1 17 m HCla 1000 m ba36.46 g b =.6 g HCl ml Br a g ml b = g Br = * 10 3 g Br
4 HEINS v4.qxd 1/30/06 1:58 PM Page (d) ( NiSO 4 )a g b = 1.31 g NiSO 4 ( HC H 3 O )a g b = 3.60 g HC H 3 O (0.75 Bi S 3 )a 514. g b = 373 g Bi S 3 (4.50 * ecules C 6 H 1 O 6 ) 6.0 * 10 3 ecules a 180. g b (e) (75 ml solutiona g ml ba0.00 g K CrO 4 b = 18 g K g solution CrO 4 = 1.35g C 6 H 1 O 6 5. Larger number of ecules: 10.0 g H O or 10.0 g H O Water has a lower ar mass than hydrogen peroxide grams of water has a lower ar mass, contains more es, and therefore more ecules than 10.0 g of H O. 6. Larger number of ecules: 5.0 g HCl or 85 g C 6 H 1 O g HCla g ba6.0 * 103 ecules b = 4.13 * 10 3 ecules HCl g C 6 H 1 O 6 a g ba6.0 * 103 ecules b HCl contains more ecules =.84 * 10 3 ecules C 6 H 1 O 6 7. Mole ratios C 3 H 7 OH + 9 O 6 CO + 8 H O 6 CO C 3 H 7 OH (d) 8 H O C 3 H 7 OH C 3 H 7 OH 9 O (e) 6 CO 8 H O 9 O 6 CO (f) 8 H O 9 O
5 HEINS v4.qxd 1/30/06 1:58 PM Page Mole ratios 3 CaCl + H 3 PO 4 Ca 3 (PO 4 ) + 6 HCl 3 CaCl 1 Ca 3 (PO 4 ) (d) 1 Ca 3 (PO 4 ) H 3 PO 4 6 HCl H 3 PO 4 (e) 6 HCl 1 Ca 3 (PO 4 ) 3 CaCl H 3 PO 4 (f) H 3 PO 4 6 HCl 9. C H 5 OH + 3 O CO + 3 H O CO C H 5 OH 1 C H 5 OH = 15.5 CO 10. The balanced equation is 4 HCl + O Cl + H O HCl Cl 4 HCl =.80 Cl 11. The balanced equation is MnO (s) + 4 HCl(aq) Cl (g) + MnCl (aq) + H O(l) 4 HCl MnO = 4.0 HCl 1 MnO 11.5 H O 1 MnCl H O = 0.65 MnCl s 1. Al 4 C H O 4 Al(OH) CH g Al 4 C 3 a g b 1 H O 1 Al 4 C 3 = 8.33 H O CH 4 4 Al(OH) 3 3 CH 4 = Al(OH)
6 HEINS v4.qxd 1/30/06 1:58 PM Page Grams of NaOH Ca(OH) + Na CO 3 NaOH + CaCO 3 The conversion is: g Ca(OH) Ca(OH) NaOH g NaOH 1500 g Ca(OH) a 1 NaOH ba ba g b = 5 * 10 g NaOH g 1 Ca(OH) 14. Grams of Zn 3 (PO 4 ) 3 Zn + H 3 PO 4 Zn 3 (PO 4 ) + 3 H The conversion is: g Zn Zn Zn 3 (PO 4 ) g Zn 3 (PO 4 ) g Zna g b 1 Zn 3(PO 4 ) a g b = 19.7 g Zn 3 Zn 3 (PO 4 ) 15. The balanced equation is Fe O C Fe + 3 CO The conversion is: kg Fe O 3 k Fe O 3 k Fe kg Fe 115 kg Fe O 3 a 1 k k Fe kg ba ba b = 87.4 kg Fe kg 1 k Fe O 3 k 16. The balanced equation is 3 Fe + 4 H O Fe 3 O H Calculate the grams of both H O and Fe to produce 375 g Fe 3 O g Fe 3 O g a 4 H O a 18.0 g b = 117 g H 1 Fe 3 O 4 O 1375 g Fe 3 O 4 a 1 3 Fe b a g b = 71 g Fe 31.6 g 1 Fe 3 O The balanced equation is C H O 4 CO + 6 H O C H 6 7 O C H 6 = 5.5 O g H Oa g b 4 CO 6 H O a g b = 13.0 g CO g C H 6 a g b 4 CO a g b =.0 * 10 g CO C H 6 (d) 1.75 of H O 4 CO 6 H O a g b = 80.7 g CO
7 HEINS v4.qxd 1/30/06 1:58 PM Page 101 (e) 15.0 C H 6 7 O C H 6 a 3.00 g O b =.80 * 10 3 g O (f) 115 g H O g C H 6 6 H O a g b = 69.5 g C H FeS + 11 O Fe O SO FeS a Fe O 3 4 FeS b = Fe O FeS 11 O 4 FeS = 1.4 O Fe O 3 8 SO Fe O 3 = 6.0 SO (d) FeS 8 SO a g b = 65.6 g SO 4 FeS (e) g SO a g b 11 O 8 SO = O (f) 11 g Fe O 3 a g b 4 FeS a 10.0 g b = 33 g FeS Fe O Hydrogen Oxygen Hydrogen is the limiting reactant
8 HEINS v4.qxd 1/30/06 1:58 PM Page 10 Hydrogen Bromine Bromine is the limiting reactant. 0. Lithium Iodine No limiting reactant. Silver Chlorine Silver is the limiting reactant. 1. Potassium Chlorine Potassium is the limiting reactant
9 HEINS v4.qxd 1/30/06 1:58 PM Page 103 Aluminum Oxygen Oxygen is the limiting reactant.. Nitrogen Oxygen Oxygen is the limiting reactant. Iron Hydrogen Oxygen Water is the limiting reactant. 3. KOH + HNO 3 KNO 3 + H O 16.0 g 1.0 g Choose one of the products and calculate its mass that would be produced from each given reactant. Using KNO 3 as the product: g KOH g a 1 KNO 3 1 KOH a g b = 8.8 g KNO g HNO 3 a g b 1 KNO 3 1 KOH a g b = 19.3 g KNO 3 Since HNO 3 produces less KNO 3, it is the limiting reactant and KOH is in excess
10 HEINS v4.qxd 1/30/06 1:58 PM Page 104 NaOH + H SO 4 Na SO 4 + H O 10.0 g 10.0 g Choose one of the products and calculate its mass that would be produced from each given reactant. Using H O as the product: g NaOHa g b H O NaOH a 18.0 g b = 4.51 g H O g H SO 4 a g b H O a 18.0 g b = 3.67 g H 1 H SO 4 O Since H SO 4 produces less H O, it is the limiting reactant and NaOH is in excess. 4. Bi(NO 3 ) H S Bi S HNO g 6.00 g Choose one of the products and calculate its mass that would be produced from each given reactant. Using Bi S 3 as the product: g Bi(NO 3 ) 3 )a g ba 1 Bi S 3 ba 514. g b = 3.5 g Bi Bi(NO 3 ) 3 S 3 (6.00 g H S)a g b 1 Bi S 3 3 H S a 514. g b = 30. g Bi S 3 Since H S produces less Bi S 3, it is the limiting reactant and Bi(NO 3 ) 3 is in excess. 3 Fe + 4 H O Fe 3 O H 40.0 g 16.0 g Choose one of the products and calculate its mass that would be produced from each given reactant. Using as the product: H (40.0 g Fe)a g b 4 H 3 Fe a.016 g b = 1.93 g H g H Oa g b 4 H 4 H O a.016 g b = 1.79 g H Since H O produces less H, it is the limiting reactant and Fe is in excess. 5. Limiting reactant calculations C 3 H O 3 CO + 4 H O Reaction between 0.0 g C 3 H 8 and 0.0 g O Convert each amount to es of CO
11 HEINS v4.qxd 1/30/06 1:58 PM Page g C 3 H 8 a g b 3 CO 1 C 3 H 8 = 1.36 es CO 10.0 g O a g b 3 CO 5 O = es CO O is the limiting reactant. The yield is es CO. Reaction between 0.0 g C 3 H 8 and 80.0 g O Convert each amount to es of CO 10.0 g C 3 H 8 a g b 3 CO 1 C 3 H 8 = 1.36 es CO g O a g b 3 CO 5 O = 1.50 es CO C 3 H 8 is the limiting reactant. The yield is 1.36 es CO. Reaction between.0 C 3 H 8 and 14.0 O According to the equation, C 3 H 8 will react with 10 O. Therefore, C 3 H 8 is the limiting reactant and 4.0 O will remain unreacted. 1.0 C 3 H 8 3 CO 1 C 3 H 8 = 6.0 CO produced (.0 C 3 H 8 ) 4 H O 1 C 3 H 8 = 8.0 H O produced When the reaction is completed, 6.0 CO, 8.0 H O, and 4.0 O will be in the container. 6. The balanced equation is C 3 H O 6 CO + 6 H O Reaction between 15.0 g C 3 H 6 and 15.0 g O. Convert each amount to es of H O g C 3 H g 6 H O C 3 H 6 = 1.07 H O g O g 6 H O O = H O The O is the limiting reactant. The yield is H O
12 HEINS v4.qxd 1/30/06 1:58 PM Page 106 Reaction between 1.0 g of C 3 H 6 and 5.0 g of O. Convert each amount to es of H O 11.0 g C 3 H g 6 H O C 3 H 6 = H O 15.0 g O g 6 H O 9 O = 0.51 H O O is the limiting reactant. The yield is 0.51 H O. Reaction between 5.0 of C 3 H 6 and 15.0 of O. Convert each to es of CO (5.0 C 3 H 6 ) 6 CO C 3 H 6 = 15 CO (15.0 O ) 6 CO = 10 CO 9 O Since is the limiting reactant. C 3 H 6 will be left unreacted. O X O 8 XO 3 The conversion is: g O O X 8 (10.0 g O a g b 1 X 8 1 O = X g X 8 = X g = 56 g> X 8 g 56 g ar mass X = = Using the periodic table we find that the element with 3.0 g> is sulfur. X + HCl XCl + H The conversion is: g H H X 1.4 g H a g b 1 X 1 H = 1.0 X 78.5 g X = 1.0 X 78.5 g = 65.4 g> 1.0 Using the periodic table we find that the element with atomic mass 65.4 is zinc
13 HEINS v4.qxd 1/30/06 1:58 PM Page Limiting reactant calculation and percentage yield Al + 3 Br AlBr 3 Reaction between 5.0 g Al and 100. g Br Calculate the grams of AlBr 3 from each reactant g Ala g b AlBr 3 a 66.7 g b = 47 g AlBr Al 3 (100. g Br )a g b AlBr 3 a 66.7 g b = 111 g AlBr 3 Br 3 Br is limiting; 111 g AlBr 3 is the theoretical yield of product. actual yield 64. g Percent yield = a b1100 = a b1100 = 57.8% theoretical yield 111 g 30. Percent yield calculation Fe(s) + CuSO 4 (aq) Cu(s) + FeSO 4 (aq) g CuSO 4 a 1 1 Cu b a g b = 159 g Cu (theoretical yield) g 1 CuSO 4 actual yield 151 g % yield = a b1100 = a b1100 = 95.0% yield of Cu theoretical yield 159 g 31. The balanced equation is 3 C + SO CS + CO Calculate the g C needed to produce 950 g CS taking into account that the yield of CS is 86.0%. First calculate the theoretical yield of CS. 950 g CS = 1.1 * 10 3 g CS 1theoretical yield Now calculate the grams of coke needed to produce 1.1 * 10 3 g CS * 10 3 g CS a g b 3 C a 1.01 g b = 5. * 10 g C 1 CS 3. The balanced equation is CaC + H O C H + Ca1OH First calculate the grams of pure CaC in the sample from the amount of C H produced C H 1 CaC a g CaC b = 34.6 g of pure CaC 1 C H CaC in the sample
14 HEINS v4.qxd 1/30/06 1:58 PM Page 108 Now calculate the percent CaC in the impure sample g CaC 44.5 g sample 1100 = 77.8% CaC in the impure sample 33. No. There are not enough screwdrivers, wrenches or pliers. 400 screwdrivers, 3600 wrenches and 100 pliers are needed for 600 tool sets. 34. A subscript is used to indicate the number of atoms in a formula. It cannot be changed without changing the identity of the substance. Coefficients are used only to balance atoms in chemical equations. They may be changed as needed to achieve a balanced equation. 35. Consider the reaction A B and assume that you have 1 gram of A. This does not guarantee that you will produce 1 gram of B because A and B have different ar masses. One gram of A does not contain the same number of ecules as 1 gram of B. However, 1 e of A does have the same number of ecules as one e of B. (Remember, 1 e = 6.0 * 10 3 ecules always.) If you determine the number of es in one gram of A and multiply by to get the number of es of B Á then from that you can determine the grams of B using its ar mass. Equations are written in terms of es not grams KO + H O + 4 CO 4 KHCO O 0.85 g CO a 1 min g b 4 KO = KO 4 CO min KO b110.0 min = 0.19 KO min 37. The conversion is: g CO min CO min O min 0.85 g CO a 1 min g b 3 O a 3.00 g 4 CO 1750 g C 6 H 1 O 6 a g b C H 5 OH a g b = 380 g C 1 C 6 H 1 O 6 H 5 OH ba g O min g O hr 60.0 min b = 8 g O 1.0 hr hr 1750 g C 6 H 1 O 6 a g b CO a g b = 370 g CO 1 C 6 H 1 O 6 Alternate Solution: 750 g C H 6 O g C H 5 OH = 370 g CO by the conservation of mass method g C H 5 OHa 1 ml 0.79 g b = 480 ml C H 5 OH
15 HEINS v4.qxd 1/30/06 1:58 PM Page CH 3 OH + 3 O CO + 4 H O The conversion is: ml CH 3 OH g CH 3 OH CH 3 OH O g O ml CH 3 OHa 0.7 g ml b g 3 O CH 3 OH a 3.00 g b = 65 g O 39. The balanced equation is The conversion is: 7 H O + N H 4 HNO H O 175 kg N H 4 a 1000 g 1 kg b g HNO 3 a 63.0 g b =.9 * 10 5 g HNO 1 N H L H O a 1000 ml 1 L b 1.41 g 1 1 ml 34.0 g a 8 H O ba 18.0 g 7 H O 175 g H O a g b 1 N H 4 a 3.05 g b = 97.6 g N 7 H O H 4 b =.1 * 10 5 g H O (d) Reaction between 750 g of N H and 15 g of H O. Convert each amount to grams of H O g N H 4 a g b 8 H O a 18.0 g b = 3.4 * 10 3 g H 1 N H 4 O 115 g H O a g b 8 H O a 18.0 g b = 75.7 g H 7 H O O 75.7 g H O can be produced. (e) Since H O is the limiting reactant, N H 4 is in excess. 115 g H O a g b 1 N H 4 a 3.05 g b = 16.8 g N 7 H O H 4 reacted 750 g N H 4 given g N H 4 used = 730 g N H 4 remaining
16 HEINS v4.qxd 1/30/06 1:58 PM Page The balanced equation is 16 HCl + KMnO 4 5 Cl + KCl + MnCl + 8 H O Reaction between 5 g KMnO 4 and 85 g HCl. Convert each to es of MnCl. 15 g KMnO 4 a 1 KMnO g KMnO 4 ba MnCl KMnO 4 b = 0.16 MnCl 185 g HCla g ba MnCl 16 HCl b = 0.9 MnCl KMnO 4 is the limiting reactant; 0.16 MnCl produced. 175 g KCla g ba8 H O KCl ba18.0 g b = 73 g H O Theoretical yield is 91 g Cl ; Percent yield: a 75 g b(100) = 8% yield 91 g (d) Reaction between 5 g HCl and 5 g KMnO 4. Convert each amount to grams of CL. (e) 1150 g HCla g ba 5 Cl 16 HCl ba70.90 g b = 91 g Cl 15 g HCla g ba 5 Cl 16 HCl ba70.90 g b = 15 g Cl 15 g KMnO 4 a g ba 5 Cl ba g b = 8 g Cl KMnO 4 HCl is the limiting; KMnO 4 is in excess; 15 g Cl will be produced. Calculate the mass of unreacted KMnO 4 : 15 g HCla g ba KMnO 4 ba g b = 14 g KMnO 16 HCl 4 will react Unreacted KMnO 4 = 5 g -14 g = 11 g KMnO 4 remain unreacted. 41. The balanced equation is 4Ag + H S + O Ag S + H O 11.1 g Aga g ba Ag S 4 Ag ba47.9 g b = 1.3 g Ag S g H Sa g ba Ag S H S ba47.9 g b = 1.0 g Ag S
17 HEINS v4.qxd 1/30/06 1:58 PM Page 111 H S is limiting 1.0 g Ag S forms g g = 0.03 grams more H S needed to completely react Ag. 4. Mass of the beaker g beaker + Ca(OH) g beaker + CaO g H O absorbed g H Oa g b =.5 * 10- H O absorbed Since the reaction is a 1:1 e, the amount of CaO in the beaker is.5 * Convert to grams. 1.5 * g CaO CaOa b = 1.6 g CaO in the beaker g -1.6 g g g O a g ba Ag S ba 47.9 g b = 1. g Ag 1 O S 11.1 g Aga g ba H S 4 Ag ba34.09 g b = 0.17 g H S reacts beaker + CaO CaO mass of the beaker 43. Pb(NO 3 ) (aq) + KI(aq) PbI (s) + KNO 3 (aq) The solid is lead (II) iodide, PbI. Double displacement reaction. Calculate the es of each reactant. [15 g Pb(NO 3 ) ] + a g b = Pb(NO 3) 115 g KIa 1 b = KI g Stoichiometric quantities of reactants are used. Theoretical yield of PbI is Actual yield: Percent yield: a g PbI a g b = PbI b1100 = 3% yield
18 HEINS v4.qxd 1/30/06 1:58 PM Page Composition of a mixture of KNO 3 and KCl. In the mixture only KCl reacts with AgNO 3. KCl(aq) + AgNO 3 (aq) AgCl(s) + KNO 3 (aq) g KCl g AgCla b =.5 g KCl in the mixture g AgCl g mixture -.5 g KCl = 7.75 g KNO 3.5 g KCl a b1100 =.5% KCl g mixture a 7.75 g KNO g mixture b1100 = 77.5% KNO The balanced equation is Zn + HCl ZnCl + H g Zn - 35 g Zn = 145 g Zn reacted with HCl 1145 g Zna g ba1 H 1 Zn b a.016 g b = 4.47 g H produced 1145 g Zna 1 HCl ba g 1 Zn ba36.46 g b = 16 g HCl reacted g Zna 1 HCl ba g 1 Zn ba36.46 g b = 01 g HCl reacts 01 g - 16 g = 39 g more HCl needed to react wih the g Zn 46. Fe(s) + CuSO 4 (aq) Cu(s) FeSO 4 (aq).0 Fe react with.0 CuSO 4 to yield.0 Cu and.0 FeSO CuSO 4 is unreacted. At the completion of the reaction, there will be.0 Cu,.0 FeSO 4, and 1.0 CuSO 4. Determine which reactant is limiting and then calculate the g FeSO 4 produced from that reactant g Fea 1 Cu ba g 1 Fe ba63.55 g b =.8 g Cu g CuSO 4 a 1 1 Cu ba ba g b = 15.9 g Cu g 1 CuSO 4 Since CuSO 4 produces less Cu, it is the limiting reactant. Determine the mass of FeSO 4 produced from 40.0 g CuSO
19 HEINS v4.qxd 1/30/06 1:58 PM Page g CuSO 4 a g b 1 FeSO 4 a g b = 38.1 g FeSO 1 CuSO 4 4 produced Calculate the mass of unreacted Fe g CuSO 4 a g b 1 Fe a g b = 14.0 g Fe will react 1 CuSO 4 Unreacted Fe = 0.0 g g = 6.0 g. Therefore, at the completion of the reaction, 15.9 g Cu, 38.1 g FeSO 4, 6.0 g Fe, and no CuSO 4 remain. 47. Limiting reactant calculation CO(g) + H (g) CH 3 OH(l) Reaction between 40.0 g CO and 10.0 g H : determine the limiting reactant by calculating the amount of CH 3 OH that would be formed from each reactant g COa g b 1 CH 3OH 1 CO a 3.04 g b = 45.8 g CH 3 OH g H a g b 1 CH 3OH a 3.04 g b = 79.5 g CH H 3 OH CO is limiting; H is in excess; 45.8 g CH 3 OH will be produced. Calculate the mass of unreacted H : g COa g b H 1 CO a.016 g b = 5.76 g H react 10.0 g H g H = 4. g H remain unreacted 48. The balanced equation is C 6 H 1 O 6 C H 5 OH + CO First calculate the theoretical yield g C 6 H 1 O 6 a g b C H 5 OH a g b 1 C 6 H 1 O 6 Then take 84.6% of the theoretical yield to obtain the actual yield. actual yield = 1theoretical yield = 3. * 10 g C H 5 OH = 3.8 * 10 g C H 5 OH (theoretical yield) = 13.8 * 10 g C H 5 OH
20 HEINS v4.qxd 1/30/06 1:58 PM Page g C H 5 OH represents 84.6% of the theoretical yield. Calculate the theoretical yield. theoretical yield = 475 g = 561 g C H 5 OH Now calculate the g C 6 H 1 O 6 needed to produce 561 g C H 5 OH g C H 5 OHa g b 1 C 6H 1 O 6 C H 5 OH a 180. g b = 1.10 * 10 3 g C 6 H 1 O The balanced equations are: CaCl (aq) + AgNO 3 (aq) Ca(NO 3 ) (aq) + AgCl(s) MgCl (aq) + AgNO 3 (aq) Mg1NO 3 (aq) + AgCl(s) 1 of each salt will produce the same amount ( ) of AgCl. MgCl has a higher percentage of Cl than CaCl because Mg has a lower atomic mass than Ca. Therefore, on an equal mass basis, MgCl will produce more AgCl than will CaCl. Calculations show that 1.00 g MgCl produces 3.01 g AgCl, and 1.00 g CaCl produces.56 g AgCl. 50. The balanced equation is Li O + H O LiOH The conversion is: g H O H O Li O g Li O kg Li O 500 g H O astronaut day a g b 1 Li O 4.1 kg Li O astronaut day 130 days13 astronauts = 3.7 * 10 kg Li O 51. The balanced equation is H SO 4 + NaCl Na SO 4 + HCl First calculate the g HCl to be produced 10.0 L HCl solutiona Then calculate the g H SO 4 required to produce the HCl * 10 4 g HCla g b 1 H SO 4 a g HCl 1 b = 1.36 * 104 g H SO 4 Finally, calculate the kg H SO 4 (96%) 1 H O a 9.88 g ba 1 kg 1000 g b = 4.1 kg Li O astronaut day 1000 ml ba 1.0 g 1 L 1.00 ml b10.40 = 1.01 * 104 g HCl * 10 4 g H SO g H SO 4 solution 0.96 g H SO 4 a 1 kg 1000 g b = 14 kg concentrated H SO
21 HEINS v4.qxd 1/30/06 1:58 PM Page Percent yield of H SO g Sa 1 b = S to start with 3.07 g S SO - 10% =.806 SO SO.806 SO 3-10% =.55 SO SO 3.55 H SO 4-10% =.73 H SO H SO 4 a g b = 3.0 g H SO 4 formed Sa 1 H SO 4 b = H 1 S SO 4 (theoretical yield).73 H SO H SO = 7.90% yield Alternate Solution: Calculation of yield. There are three chemical steps to the formation of H SO 4. Each step has a 10% loss of yield. Step 1: Step : Step 3: 100% yield - 10% = 90.00% yield 90.00% yield - 10% = 81.00% yield 81.00% yield - 10% = 7.90% yield Now calculate the grams of product. One e of sulfur will yield a maximum of 1 H SO 4. Therefore S will give a maximum of H SO Sa 1 H SO 4 1 S ba g b = 3.0 g H SO 4 yield 53. According to the equations, the es of CO come from both reactions and the es H O come from only the first reaction. So the NaHCO 3 = * H O = * = NaHCO NaHCO 3 a g b = 6.00 g NaHCO 3 in the sample g NaHCO 3 + Na CO g NaHCO 3 = 4.00 g Na CO 3 in the sample
22 HEINS v4.qxd 1/30/06 1:58 PM Page 116 a 6.00 g NaHCO 3 b1100 = 60.0% NaHCO g 3 a 4.00 g Na CO 3 b1100 = 40.0% Na g CO The balanced equation is KClO 3 KCl + 3 O 1.8 g mixture g residue = 3.37 g O lost by heating Because the O lost came only from KClO 3, we can use it to calculate the amount of KClO 3 in the mixture. The conversion is: g O O KClO 3 g KClO g O a g ba KClO 3 ba 1.6 g b = 8.61 g KClO 3 O 3 in the mixture a 8.61 g KClO g sample b1100 = 67.% KClO The balanced equation is Al(OH) 3 (s) + 3 HCl(aq) AlCl 3 (aq) + 3 H O(l) The conversion is: L HCl g HCl HCl Al(OH) 3 : g Al(OH) 3 a.5 L day g HCl ba3.0 ba 1 L g b 1 Al1OH 3 3 HCl Now calculate the number of 400. mg tablets that can be made from 5.3 g Al(OH) g Al(OH) mg a ba 1 tablet b = 13 tablets>day day g 400. mg a g b = 5.3 g Al1OH 3 >day P + 5 O P 4 O 10 P 4 O H O 4 H 3 PO 4 In the first reaction: 10.0 g Pa 1 b = P g g O a g b = O P 3.44 P This is a ratio of = O 5.00 O Therefore, P is the limiting reactant and the P 4 O 10 produced is:
23 HEINS v4.qxd 1/30/06 1:58 PM Page P 1 P 4O 10 4 P = 0.16 P 4O 10 In the second reaction: g H Oa g b = 0.83 H O H O and we have 0.16 P 4 O 10. The ratio of is = P 4 O Therefore, H O is the limiting reactant and the H 3 PO 4 produced is: H O 4 H 3PO 4 6 H O a g b = 54.4 g H 3 PO
Steps for balancing a chemical equation
The Chemical Equation: A Chemical Recipe Dr. Gergens - SD Mesa College A. Learn the meaning of these arrows. B. The chemical equation is the shorthand notation for a chemical reaction. A chemical equation
More informationAnswers and Solutions to Text Problems
Chapter 7 Answers and Solutions 7 Answers and Solutions to Text Problems 7.1 A mole is the amount of a substance that contains 6.02 x 10 23 items. For example, one mole of water contains 6.02 10 23 molecules
More informationUnit 10A Stoichiometry Notes
Unit 10A Stoichiometry Notes Stoichiometry is a big word for a process that chemist s use to calculate amounts in reactions. It makes use of the coefficient ratio set up by balanced reaction equations
More informationChapter 5. Chemical Reactions and Equations. Introduction. Chapter 5 Topics. 5.1 What is a Chemical Reaction
Introduction Chapter 5 Chemical Reactions and Equations Chemical reactions occur all around us. How do we make sense of these changes? What patterns can we find? 1 2 Copyright The McGraw-Hill Companies,
More information1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) 3 (s) + H 2 (g)
1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) A) 1 B) 2 C) 4 D) 5 E) Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) Al (s) + H 2 O (l)? Al(OH)
More informationMoles. Moles. Moles. Moles. Balancing Eqns. Balancing. Balancing Eqns. Symbols Yields or Produces. Like a recipe:
Like a recipe: Balancing Eqns Reactants Products 2H 2 (g) + O 2 (g) 2H 2 O(l) coefficients subscripts Balancing Eqns Balancing Symbols (s) (l) (aq) (g) or Yields or Produces solid liquid (pure liquid)
More informationBalancing Chemical Equations Worksheet
Balancing Chemical Equations Worksheet Student Instructions 1. Identify the reactants and products and write a word equation. 2. Write the correct chemical formula for each of the reactants and the products.
More informationStoichiometry Review
Stoichiometry Review There are 20 problems in this review set. Answers, including problem set-up, can be found in the second half of this document. 1. N 2 (g) + 3H 2 (g) --------> 2NH 3 (g) a. nitrogen
More informationMoles. Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations
Moles Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations Micro World atoms & molecules Macro World grams Atomic mass is the mass of an
More informationCHEMICAL REACTIONS. Chemistry 51 Chapter 6
CHEMICAL REACTIONS A chemical reaction is a rearrangement of atoms in which some of the original bonds are broken and new bonds are formed to give different chemical structures. In a chemical reaction,
More informationCHEMISTRY COMPUTING FORMULA MASS WORKSHEET
CHEMISTRY COMPUTING FORMULA MASS WORKSHEET Directions: Find the formula mass of the following compounds. Round atomic masses to the tenth of a decimal place. Place your final answer in the FORMULA MASS
More informationChapter 11. Electrochemistry Oxidation and Reduction Reactions. Oxidation-Reduction Reactions. Oxidation-Reduction Reactions
Oxidation-Reduction Reactions Chapter 11 Electrochemistry Oxidation and Reduction Reactions An oxidation and reduction reaction occurs in both aqueous solutions and in reactions where substances are burned
More information1. Read P. 368-375, P. 382-387 & P. 429-436; P. 375 # 1-11 & P. 389 # 1,7,9,12,15; P. 436 #1, 7, 8, 11
SCH3U- R.H.KING ACADEMY SOLUTION & ACID/BASE WORKSHEET Name: The importance of water - MAKING CONNECTION READING 1. Read P. 368-375, P. 382-387 & P. 429-436; P. 375 # 1-11 & P. 389 # 1,7,9,12,15; P. 436
More informationUNIT (4) CALCULATIONS AND CHEMICAL REACTIONS
UNIT (4) CALCULATIONS AND CHEMICAL REACTIONS 4.1 Formula Masses Recall that the decimal number written under the symbol of the element in the periodic table is the atomic mass of the element. 1 7 8 12
More informationstoichiometry = the numerical relationships between chemical amounts in a reaction.
1 REACTIONS AND YIELD ANSWERS stoichiometry = the numerical relationships between chemical amounts in a reaction. 2C 8 H 18 (l) + 25O 2 16CO 2 (g) + 18H 2 O(g) From the equation, 16 moles of CO 2 (a greenhouse
More informationChapter 5 Chemical Quantities and Reactions. Collection Terms. 5.1 The Mole. A Mole of a Compound. A Mole of Atoms.
Chapter 5 Chemical Quantities and Reactions 5.1 The Mole Collection Terms A collection term states a specific number of items. 1 dozen donuts = 12 donuts 1 ream of paper = 500 sheets 1 case = 24 cans 1
More informationChemical Reactions 2 The Chemical Equation
Chemical Reactions 2 The Chemical Equation INFORMATION Chemical equations are symbolic devices used to represent actual chemical reactions. The left side of the equation, called the reactants, is separated
More informationChemical Equations and Chemical Reactions. Chapter 8.1
Chemical Equations and Chemical Reactions Chapter 8.1 Objectives List observations that suggest that a chemical reaction has taken place List the requirements for a correctly written chemical equation.
More information= 11.0 g (assuming 100 washers is exact).
CHAPTER 8 1. 100 washers 0.110 g 1 washer 100. g 1 washer 0.110 g = 11.0 g (assuming 100 washers is exact). = 909 washers 2. The empirical formula is CFH from the structure given. The empirical formula
More informationTutorial 4 SOLUTION STOICHIOMETRY. Solution stoichiometry calculations involve chemical reactions taking place in solution.
T-27 Tutorial 4 SOLUTION STOICHIOMETRY Solution stoichiometry calculations involve chemical reactions taking place in solution. Of the various methods of expressing solution concentration the most convenient
More informationChapter 6 Notes Science 10 Name:
6.1 Types of Chemical Reactions a) Synthesis (A + B AB) Synthesis reactions are also known as reactions. When this occurs two or more reactants (usually elements) join to form a. A + B AB, where A and
More informationWriting, Balancing and Predicting Products of Chemical Reactions.
Writing, Balancing and Predicting Products of Chemical Reactions. A chemical equation is a concise shorthand expression which represents the relative amount of reactants and products involved in a chemical
More informationChemistry: Chemical Equations
Chemistry: Chemical Equations Write a balanced chemical equation for each word equation. Include the phase of each substance in the equation. Classify the reaction as synthesis, decomposition, single replacement,
More informationBalancing Chemical Equations Practice
Science Objectives Students will describe what reactants and products in a chemical equation mean. Students will explain the difference between coefficients and subscripts in chemical equations. Students
More informationPART I: MULTIPLE CHOICE (30 multiple choice questions. Each multiple choice question is worth 2 points)
CHEMISTRY 123-07 Midterm #1 Answer key October 14, 2010 Statistics: Average: 74 p (74%); Highest: 97 p (95%); Lowest: 33 p (33%) Number of students performing at or above average: 67 (57%) Number of students
More informationChapter 3 Mass Relationships in Chemical Reactions
Chapter 3 Mass Relationships in Chemical Reactions Student: 1. An atom of bromine has a mass about four times greater than that of an atom of neon. Which choice makes the correct comparison of the relative
More informationSolution a homogeneous mixture = A solvent + solute(s) Aqueous solution water is the solvent
Solution a homogeneous mixture = A solvent + solute(s) Aqueous solution water is the solvent Water a polar solvent: dissolves most ionic compounds as well as many molecular compounds Aqueous solution:
More informationUnit 9 Stoichiometry Notes (The Mole Continues)
Unit 9 Stoichiometry Notes (The Mole Continues) is a big word for a process that chemist s use to calculate amounts in reactions. It makes use of the coefficient ratio set up by balanced reaction equations
More informationChapter 8: Chemical Equations and Reactions
Chapter 8: Chemical Equations and Reactions I. Describing Chemical Reactions A. A chemical reaction is the process by which one or more substances are changed into one or more different substances. A chemical
More informationAppendix D. Reaction Stoichiometry D.1 INTRODUCTION
Appendix D Reaction Stoichiometry D.1 INTRODUCTION In Appendix A, the stoichiometry of elements and compounds was presented. There, the relationships among grams, moles and number of atoms and molecules
More informationChapter 8 - Chemical Equations and Reactions
Chapter 8 - Chemical Equations and Reactions 8-1 Describing Chemical Reactions I. Introduction A. Reactants 1. Original substances entering into a chemical rxn B. Products 1. The resulting substances from
More informationFormulas, Equations and Moles
Chapter 3 Formulas, Equations and Moles Interpreting Chemical Equations You can interpret a balanced chemical equation in many ways. On a microscopic level, two molecules of H 2 react with one molecule
More informationChemical Equations. Chemical Equations. Chemical reactions describe processes involving chemical change
Chemical Reactions Chemical Equations Chemical reactions describe processes involving chemical change The chemical change involves rearranging matter Converting one or more pure substances into new pure
More informationChemistry B11 Chapter 4 Chemical reactions
Chemistry B11 Chapter 4 Chemical reactions Chemical reactions are classified into five groups: A + B AB Synthesis reactions (Combination) H + O H O AB A + B Decomposition reactions (Analysis) NaCl Na +Cl
More informationSample Exercise 3.1 Interpreting and Balancing Chemical Equations
Sample Exercise 3.1 Interpreting and Balancing Chemical Equations The following diagram represents a chemical reaction in which the red spheres are oxygen atoms and the blue spheres are nitrogen atoms.
More informationName: Class: Date: 2 4 (aq)
Name: Class: Date: Unit 4 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1) The balanced molecular equation for complete neutralization of
More informationSCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, "Chemistry 11", McGraw-Hill Ryerson, 2001
SCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, "Chemistry 11", McGraw-Hill Ryerson, 2001 1. A small pin contains 0.0178 mol of iron. How many atoms of iron are in the pin? 2. A sample
More informationW1 WORKSHOP ON STOICHIOMETRY
INTRODUCTION W1 WORKSHOP ON STOICHIOMETRY These notes and exercises are designed to introduce you to the basic concepts required to understand a chemical formula or equation. Relative atomic masses of
More informationCHEM 110: CHAPTER 3: STOICHIOMETRY: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS
1 CHEM 110: CHAPTER 3: STOICHIOMETRY: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS The Chemical Equation A chemical equation concisely shows the initial (reactants) and final (products) results of
More informationChapter 3: Stoichiometry
Chapter 3: Stoichiometry Key Skills: Balance chemical equations Predict the products of simple combination, decomposition, and combustion reactions. Calculate formula weights Convert grams to moles and
More informationProblem Solving. Stoichiometry of Gases
Skills Worksheet Problem Solving Stoichiometry of Gases Now that you have worked with relationships among moles, mass, and volumes of gases, you can easily put these to work in stoichiometry calculations.
More informationStoichiometry. Unit Outline
3 Stoichiometry Unit Outline 3.1 The Mole and Molar Mass 3.2 Stoichiometry and Compound Formulas 3.3 Stoichiometry and Chemical Reactions 3.4 Stoichiometry and Limiting Reactants 3.5 Chemical Analysis
More informationName Class Date. Section: Calculating Quantities in Reactions. Complete each statement below by writing the correct term or phrase.
Skills Worksheet Concept Review Section: Calculating Quantities in Reactions Complete each statement below by writing the correct term or phrase. 1. All stoichiometric calculations involving equations
More informationAqueous Solutions. Water is the dissolving medium, or solvent. Some Properties of Water. A Solute. Types of Chemical Reactions.
Aqueous Solutions and Solution Stoichiometry Water is the dissolving medium, or solvent. Some Properties of Water Water is bent or V-shaped. The O-H bonds are covalent. Water is a polar molecule. Hydration
More informationIB Chemistry. DP Chemistry Review
DP Chemistry Review Topic 1: Quantitative chemistry 1.1 The mole concept and Avogadro s constant Assessment statement Apply the mole concept to substances. Determine the number of particles and the amount
More informationWord Equations and Balancing Equations. Video Notes
Word Equations and Balancing Equations Video Notes In this lesson, you will: Use the law of conservation of mass and provide standard rules for writing and balancing equations. Write and balance equations
More information4. Balanced chemical equations tell us in what molar ratios substances combine to form products, not in what mass proportions they combine.
CHAPTER 9 1. The coefficients of the balanced chemical equation for a reaction give the relative numbers of molecules of reactants and products that are involved in the reaction.. The coefficients of the
More informationWriting and Balancing Chemical Equations
Name Writing and Balancing Chemical Equations Period When a substance undergoes a chemical reaction, chemical bonds are broken and new bonds are formed. This results in one or more new substances, often
More informationNET IONIC EQUATIONS. A balanced chemical equation can describe all chemical reactions, an example of such an equation is:
NET IONIC EQUATIONS A balanced chemical equation can describe all chemical reactions, an example of such an equation is: NaCl + AgNO 3 AgCl + NaNO 3 In this case, the simple formulas of the various reactants
More informationCalculations and Chemical Equations. Example: Hydrogen atomic weight = 1.008 amu Carbon atomic weight = 12.001 amu
Calculations and Chemical Equations Atomic mass: Mass of an atom of an element, expressed in atomic mass units Atomic mass unit (amu): 1.661 x 10-24 g Atomic weight: Average mass of all isotopes of a given
More informationChapter 1 The Atomic Nature of Matter
Chapter 1 The Atomic Nature of Matter 6. Substances that cannot be decomposed into two or more simpler substances by chemical means are called a. pure substances. b. compounds. c. molecules. d. elements.
More informationChemistry Themed. Types of Reactions
Chemistry Themed Types of Reactions 1 2 Chemistry in the Community-2015-2016 Types of Reactions Date In-Class Assignment Homework T 10/20 TEST on Reactivity of Metals and Redox None W 10/21 Late Start
More informationTypes of Reactions. CHM 130LL: Chemical Reactions. Introduction. General Information
Introduction CHM 130LL: Chemical Reactions We often study chemistry to understand how and why chemicals (reactants) can be transformed into different chemicals (products) via a chemical reaction: Reactants
More informationLiquid phase. Balance equation Moles A Stoic. coefficient. Aqueous phase
STOICHIOMETRY Objective The purpose of this exercise is to give you some practice on some Stoichiometry calculations. Discussion The molecular mass of a compound is the sum of the atomic masses of all
More informationCalculating Atoms, Ions, or Molecules Using Moles
TEKS REVIEW 8B Calculating Atoms, Ions, or Molecules Using Moles TEKS 8B READINESS Use the mole concept to calculate the number of atoms, ions, or molecules in a sample TEKS_TXT of material. Vocabulary
More informationAtomic Structure. Name Mass Charge Location Protons 1 +1 Nucleus Neutrons 1 0 Nucleus Electrons 1/1837-1 Orbit nucleus in outer shells
Atomic Structure called nucleons Name Mass Charge Location Protons 1 +1 Nucleus Neutrons 1 0 Nucleus Electrons 1/1837-1 Orbit nucleus in outer shells The number of protons equals the atomic number This
More informationStoichiometry. 1. The total number of moles represented by 20 grams of calcium carbonate is (1) 1; (2) 2; (3) 0.1; (4) 0.2.
Stoichiometry 1 The total number of moles represented by 20 grams of calcium carbonate is (1) 1; (2) 2; (3) 01; (4) 02 2 A 44 gram sample of a hydrate was heated until the water of hydration was driven
More informationDepartment of Chemical Engineering Review Sheet Chemical Reactions Prepared by Dr. Timothy D. Placek from various sources
Department of Chemical Engineering Review Sheet Chemical Reactions Prepared by Dr. Timothy D. Placek from various sources Introduction This document is intended to help you review the basics of writing
More informationReactions in Aqueous Solution
CHAPTER 7 1. Water is the most universal of all liquids. Water has a relatively large heat capacity and a relatively large liquid range, which means it can absorb the heat liberated by many reactions while
More informationPercent Composition and Molecular Formula Worksheet
Percent Composition and Molecular Formula Worksheet 1. What s the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% 2. If the molar mass of the compound in problem 1 is
More informationChem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses
Chem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses B. Calculations of moles C. Calculations of number of atoms from moles/molar masses 1. Avagadro
More information6 Reactions in Aqueous Solutions
6 Reactions in Aqueous Solutions Water is by far the most common medium in which chemical reactions occur naturally. It is not hard to see this: 70% of our body mass is water and about 70% of the surface
More informationAPPENDIX B: EXERCISES
BUILDING CHEMISTRY LABORATORY SESSIONS APPENDIX B: EXERCISES Molecular mass, the mole, and mass percent Relative atomic and molecular mass Relative atomic mass (A r ) is a constant that expresses the ratio
More informationChapter 3. Chemical Reactions and Reaction Stoichiometry. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT
Lecture Presentation Chapter 3 Chemical Reactions and Reaction James F. Kirby Quinnipiac University Hamden, CT The study of the mass relationships in chemistry Based on the Law of Conservation of Mass
More informationChemistry 51 Chapter 8 TYPES OF SOLUTIONS. A solution is a homogeneous mixture of two substances: a solute and a solvent.
TYPES OF SOLUTIONS A solution is a homogeneous mixture of two substances: a solute and a solvent. Solute: substance being dissolved; present in lesser amount. Solvent: substance doing the dissolving; present
More informationStoichiometry and Aqueous Reactions (Chapter 4)
Stoichiometry and Aqueous Reactions (Chapter 4) Chemical Equations 1. Balancing Chemical Equations (from Chapter 3) Adjust coefficients to get equal numbers of each kind of element on both sides of arrow.
More informationDavid A. Katz Chemist, Educator, Science Communicator, and Consultant Department of Chemistry, Pima Community College
WRITING CHEMICAL EQUATIONS 2004, 2002, 1989 by David A. Katz. All rights reserved. Permission for classroom used provided original copyright is included. David A. Katz Chemist, Educator, Science Communicator,
More informationLimiting Reagent Worksheet #1
Limiting Reagent Worksheet #1 1. Given the following reaction: (Balance the equation first!) C 3 H 8 + O 2 -------> CO 2 + H 2 O a) If you start with 14.8 g of C 3 H 8 and 3.44 g of O 2, determine the
More informationChapter 6 Chemical Calculations
Chapter 6 Chemical Calculations 1 Submicroscopic Macroscopic 2 Chapter Outline 1. Formula Masses (Ch 6.1) 2. Percent Composition (supplemental material) 3. The Mole & Avogadro s Number (Ch 6.2) 4. Molar
More informationI N V E S T I C E D O R O Z V O J E V Z D Ě L Á V Á N Í CHEMICAL REACTIONS
Chemical reaction = process during which original substances change to new substances, reactants turn to... The bonds of reactants... and new bonds are... The classification of reactions: 1. Classification
More informationChapter 7: Chemical Reactions
Chapter 7 Page 1 Chapter 7: Chemical Reactions A chemical reaction: a process in which at least one new substance is formed as the result of a chemical change. A + B C + D Reactants Products Evidence that
More informationEnglish already has many collective nouns for fixed, given numbers of objects. Some of the more common collective nouns are shown in Table 7.1.
96 Chapter 7: Calculations with Chemical Formulas and Chemical Reactions Chemical reactions are written showing a few individual atoms or molecules reacting to form a few atoms or molecules of products.
More information1. What is the molecular formula of a compound with the empirical formula PO and a gram-molecular mass of 284 grams?
Name: Tuesday, May 20, 2008 1. What is the molecular formula of a compound with the empirical formula PO and a gram-molecular mass of 284 grams? 2 5 1. P2O 5 3. P10O4 2. P5O 2 4. P4O10 2. Which substance
More informationCP Chemistry Review for Stoichiometry Test
CP Chemistry Review for Stoichiometry Test Stoichiometry Problems (one given reactant): 1. Make sure you have a balanced chemical equation 2. Convert to moles of the known substance. (Use the periodic
More informationBalancing Chemical Equations Worksheet Intermediate Level
Balancing Chemical Equations Worksheet Intermediate Level Neutralization Reactions Salts are produced by the action of acids. Salts are written metal first, then non-metal. Eg. NaCl not ClNa Acid + Base
More informationExperiment 5. Chemical Reactions A + X AX AX A + X A + BX AX + B AZ + BX AX + BZ
Experiment 5 Chemical Reactions OBJECTIVES 1. To observe the various criteria that are used to indicate that a chemical reaction has occurred. 2. To convert word equations into balanced inorganic chemical
More information4 theoretical problems 2 practical problems
1 st 4 theoretical problems 2 practical problems FIRST INTERNATIONAL CHEMISTRY OLYMPIAD PRAGUE 1968 CZECHOSLOVAKIA THEORETICAL PROBLEMS PROBLEM 1 A mixture of hydrogen and chlorine kept in a closed flask
More informationRedox and Electrochemistry
Name: Thursday, May 08, 2008 Redox and Electrochemistry 1. A diagram of a chemical cell and an equation are shown below. When the switch is closed, electrons will flow from 1. the Pb(s) to the Cu(s) 2+
More informationFormulae, stoichiometry and the mole concept
3 Formulae, stoichiometry and the mole concept Content 3.1 Symbols, Formulae and Chemical equations 3.2 Concept of Relative Mass 3.3 Mole Concept and Stoichiometry Learning Outcomes Candidates should be
More informationBalance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O
Balance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O Ans: 8 KClO 3 + C 12 H 22 O 11 8 KCl + 12 CO 2 + 11 H 2 O 3.2 Chemical Symbols at Different levels Chemical symbols represent
More informationMolar Mass Worksheet Answer Key
Molar Mass Worksheet Answer Key Calculate the molar masses of the following chemicals: 1) Cl 2 71 g/mol 2) KOH 56.1 g/mol 3) BeCl 2 80 g/mol 4) FeCl 3 162.3 g/mol 5) BF 3 67.8 g/mol 6) CCl 2 F 2 121 g/mol
More informationSTOICHIOMETRY UNIT 1 LEARNING OUTCOMES. At the end of this unit students will be expected to:
STOICHIOMETRY LEARNING OUTCOMES At the end of this unit students will be expected to: UNIT 1 THE MOLE AND MOLAR MASS define molar mass and perform mole-mass inter-conversions for pure substances explain
More informationCLASS TEST GRADE 11. PHYSICAL SCIENCES: CHEMISTRY Test 6: Chemical change
CLASS TEST GRADE PHYSICAL SCIENCES: CHEMISTRY Test 6: Chemical change MARKS: 45 TIME: hour INSTRUCTIONS AND INFORMATION. Answer ALL the questions. 2. You may use non-programmable calculators. 3. You may
More informationMASS RELATIONSHIPS IN CHEMICAL REACTIONS
MASS RELATIONSHIPS IN CHEMICAL REACTIONS 1. The mole, Avogadro s number and molar mass of an element. Molecular mass (molecular weight) 3. Percent composition of compounds 4. Empirical and Molecular formulas
More informationMoles, Molecules, and Grams Worksheet Answer Key
Moles, Molecules, and Grams Worksheet Answer Key 1) How many are there in 24 grams of FeF 3? 1.28 x 10 23 2) How many are there in 450 grams of Na 2 SO 4? 1.91 x 10 24 3) How many grams are there in 2.3
More informationChapter 3! Stoichiometry: Calculations with Chemical Formulas and Equations. Stoichiometry
Chapter 3! : Calculations with Chemical Formulas and Equations Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2 H 2 O (g) Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2
More informationCalculation of Molar Masses. Molar Mass. Solutions. Solutions
Molar Mass Molar mass = Mass in grams of one mole of any element, numerically equal to its atomic weight Molar mass of molecules can be determined from the chemical formula and molar masses of elements
More informationChemistry Final Study Guide
Name: Class: Date: Chemistry Final Study Guide Multiple Choice Identify the choice that best completes the statement or answers the question. 1. The electrons involved in the formation of a covalent bond
More information2. The percent yield is the maximum amount of product that can be produced from the given amount of limiting reactant.
UNIT 6 stoichiometry practice test True/False Indicate whether the statement is true or false. moles F 1. The mole ratio is a comparison of how many grams of one substance are required to participate in
More informationStoichiometry CHAPTER 12
CHAPTER 1 Stoichiometry What You ll Learn You will write mole ratios from balanced chemical equations. You will calculate the number of moles and the mass of a reactant or product when given the number
More informationEDEXCEL INTERNATIONAL GCSE CHEMISTRY EDEXCEL CERTIFICATE IN CHEMISTRY ANSWERS SECTION E
EDEXCEL INTERNATIONAL GCSE CHEMISTRY EDEXCEL CERTIFICATE IN CHEMISTRY ANSWERS SECTION E (To save endless repetition, wherever they are included, comments are intended for homeschooling parents who may
More informationChapter 5, Calculations and the Chemical Equation
1. How many iron atoms are present in one mole of iron? Ans. 6.02 1023 atoms 2. How many grams of sulfur are found in 0.150 mol of sulfur? [Use atomic weight: S, 32.06 amu] Ans. 4.81 g 3. How many moles
More informationneutrons are present?
AP Chem Summer Assignment Worksheet #1 Atomic Structure 1. a) For the ion 39 K +, state how many electrons, how many protons, and how many 19 neutrons are present? b) Which of these particles has the smallest
More informationStoichiometry. What is the atomic mass for carbon? For zinc?
Stoichiometry Atomic Mass (atomic weight) Atoms are so small, it is difficult to discuss how much they weigh in grams We use atomic mass units an atomic mass unit (AMU) is one twelfth the mass of the catbon-12
More information2. Write the chemical formula(s) of the product(s) and balance the following spontaneous reactions.
1. Using the Activity Series on the Useful Information pages of the exam write the chemical formula(s) of the product(s) and balance the following reactions. Identify all products phases as either (g)as,
More informationREVIEW QUESTIONS Chapter 8
Chemistry 51 ANSWER KEY REVIEW QUESTIONS Chapter 8 1. Identify each of the diagrams below as strong electrolyte, weak electrolyte or non-electrolyte: (a) Non-electrolyte (no ions present) (b) Weak electrolyte
More informationChemistry Ch 15 (Solutions) Study Guide Introduction
Chemistry Ch 15 (Solutions) Study Guide Introduction Name: Note: a word marked (?) is a vocabulary word you should know the meaning of. A homogeneous (?) mixture, or, is a mixture in which the individual
More informationName Date Class STOICHIOMETRY. SECTION 12.1 THE ARITHMETIC OF EQUATIONS (pages 353 358)
Name Date Class 1 STOICHIOMETRY SECTION 1.1 THE ARITHMETIC OF EQUATIONS (pages 353 358) This section explains how to calculate the amount of reactants required or product formed in a nonchemical process.
More information1332 CHAPTER 18 Sample Questions
1332 CHAPTER 18 Sample Questions Couple E 0 Couple E 0 Br 2 (l) + 2e 2Br (aq) +1.06 V AuCl 4 + 3e Au + 4Cl +1.00 V Ag + + e Ag +0.80 V Hg 2+ 2 + 2e 2 Hg +0.79 V Fe 3+ (aq) + e Fe 2+ (aq) +0.77 V Cu 2+
More information2. DECOMPOSITION REACTION ( A couple have a heated argument and break up )
TYPES OF CHEMICAL REACTIONS Most reactions can be classified into one of five categories by examining the types of reactants and products involved in the reaction. Knowing the types of reactions can help
More information