Chem 305. Quiz 4. Chapters 5 and 6 Dr. Alino Spring VERSION A. Name: Date:

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1 Chem 305. Quiz 4. Chapters 5 and 6 Dr. Alino Spring 2015 VERSION A. Name: Date: I. Multiple Choice. 1. The following reactions matches the type of reaction, except? DEFINITION a. 2Fe + 3Cl 2 2FeCl 3 ; combination b. 2 BaO 2 2 BaO + O 2 ; decomposition c. Zn + 2HCl ZnCl 2 + H 2 ; single replacement d. Sb 2 S 3 + 6HCl 2SbCl 3 + 3H 2 S ; double replacement e. 4Na + O 2 2Na 2 O ;combustion 2. Which of the following compounds is one of the products of the reaction: Predicting Products CaCl KF CaF 2(s) + 2 KCl Ca 2+ Cl K + F Ca 2+ F K + Cl CaCl 2 + KF + a. CaF c. Cl 2 F b. KCl 2 d. CaF 2 3. Which of the following compounds cannot be in its ionized form in aqueous solutions? SOLUBILITY a. CdCl 2 c. Al(ClO 3 ) 3 b. HgNO 3 d. ZnSO 4 e. none, all compounds will ionize in water 4. The following statements about net ionic equations are true except. CONCEPT a. It tells you which species has changed into ions in solution and which species experienced the opposite process. b. It shows you which species are reacting c. It shows you the spectator ions d. It tells you how various species exists, whether as ions in solutions or as undissociated molecules 5. For the balanced reaction: TOTAL IONIC EQUATION 3Co(NO 3 ) 2 + Fe 2 (SO 4 ) 3 2 Fe(NO 3 ) 3 + 3CoSO 4 The total ionic equation is a. 3 Co NO 3 + 2Fe SO 4 2 Fe NO 3 b. 3 Co NO 3 + 2Fe SO 4 2 Fe NO 3 c. 3 Co NO 3 + Fe SO 4 2 Fe NO 3 d. 3 Co NO 3 + 2Fe SO 4 2 Fe NO 3 1

2 6. When a single molecule of calcium sulfate is added to a beaker filled with water, the ions found in the beaker are: SOLUBILITY a. 1 Ca 2+ and 1 SO 4 ions c. 2 Ca 2+ and 2 SO 4 ions b. 1 Ca 2+ and 1 SO 3 ions d. none, the molecule is insoluble to water 7. When a single molecule rubidium phosphate is added to a beaker filled with water, the ions found in the beaker are: SOLUBILITY a. 3 Rb + 3 and 1 PO 3 ions c. 3 Rb + 3 and 1 PO 4 ions b. 1 Rb + 3 and 3 PO 4 ions d. none, the molecule is insoluble to water 8. Predict the products PREDICTING THE PRODUCTS CuO (s) + 2HCl CuCl 2 + H 2 O (l) Copper (II) oxide + hydrochloric acid + (Name; formula multiple choice) a. Cu(OCl) 2 and H 2 c. H 2 O and CuCl 2 b. Cu(OH) 2 and Cl 2 d. HOCl and Cu 9. Balance the equation. BALANCING THE EQUATION C 2 H 6(g) + O 2(g) CO 2(g) + H 2 O (l) a. C 2 H 6(g) + 3 O 2(g) 2 CO 2(g) + 3 H 2 O (l) b. C 2 H 6(g) + 6 O 2(g) 2 CO 2(g) + 6 H 2 O (l) c. C 2 H 6(g) + 7 O 2(g) 2 CO 2(g) + 3 H 2 O (l) d. 2 C 2 H 6(g) + 7 O 2(g) 4 CO 2(g) + 6 H 2 O (l) e. None of the above What is the oxidation state for P in PO 4 3 PO 4 = 3 P + 4(2) = 3 P = P = + 5 a. 3 c. 5 b. + 5 d. none of the above? OXIDATION STATE 2

3 11. What is the oxidation state for As in H 3 AsO 4? OXIDATION STATE H 3 AsO 4 = 0 3(+1) + As + 4(2)= 0 As = As = + 5 a. 3 c. 5 b. + 5 d. none of the above 12. Which of the reactant is the oxidizing agent? REDOX; ONE THAT IS REDUCED H = +1 Cl=1 Fe=0 H=0 Fe=+2 Cl=1 2 HCl + Fe H2 + FeCl 2 a. HCl b. Fe 13. Given the following metal and compound combination, how many of these will NOT have a reaction? ACTIVITY SERIES i. K + BaCl 2 ii. Cr + CuSO 3 iii. Fe + NiCl 2 iv. Sn + NiBr 2 a. 0 c. 2 b. 1 d. 3 e. all Which of the reactant is oxidized? REDOX CoCl Ag Co + 2 AgCl a. Ag b. CoCl How many phosphate atoms are there in 20.5 grams of Mg 3 (PO 4 ) 2?COMPUTATION MOLECULES TO MASS MM Mg 3 (PO 4 ) 2 = g/mol 20.5 g Mg 3 (PO 4 ) 2 x 1 mol Mg 3 (PO 4 ) 2 x 2 mol P x x P atoms = 9.39 x10 22 P atoms g Mg 3 (PO 4 ) 2 1 mol Mg 3 (PO 4 ) 2 1 mol P a. 7.35x10 22 P atoms c. 8.29x10 27 P atoms b x P atoms d x P atoms 16. How many moles of calcium atoms are present in 108 grams of calcium nitrate? MM Ca(NO 3 ) 2 = g/mol? mol Ca atoms = 108 g Ca(NO 3 ) 2 x 1 mol Ca(NO 3 ) 2 x 1 mol Ca atoms = mol Ca atoms g Ca(NO 3 ) 2 1 mol Ca(NO 3 ) 2 a mol Ca atoms c x 10 4 mol Ca atoms b x mol Ca atoms d x mol Ca atoms 3

4 17. How many grams of potassium hydroxide are required to neutralize 4.86 g of sulfuric acid? Acid Base Neutralization 2 KOH + H 2 SO 4 2 H 2 O + K 2 SO g/mol g/mol g/mol g/mol?gkoh = 4.86 g H 2 SO 4 x 1 mol H 2 SO 4 x 2 mol KOH x g KOH = g H 2 SO 4 1 mol H 2 SO 4 1 mol KOH a g KOH c g KOH b g KOH d g KOH 18. How many grams of the second product (NH 4 Cl) are produced should the following reaction produce 26.5 grams of CdS? CdCl 2 + (NH 4 ) 2 S CdS + 2 NH 4 Cl g/mol g/mol g/mol g/mol? g NH 4 Cl = 26.5 g CdS x 1 mol CdS x 2 mol NH 4 Cl x g NH 4 Cl = 19.6 g NH 4 Cl g CdS 1 mol CdS 1 mol NH 4 Cl a g NH 4 Cl c g NH 4 Cl b g NH 4 Cl d g NH 4 Cl 19. If a mixture containing 1.8 moles of HBr and 0.50 moles of Ba(OH) 2 is allowed to react according to the equation: LIMITING REAGENT PROBLEM 2 HBr + Ba(OH) 2 2 H 2 O + BaBr g/mol g/mol g/mol g/mol? mol BaBr 2 = 1.8 mol HBr x 1 mol BaBr 2 = 0.9 mol BaBr 2 2 mol HBr? mol BaBr 2 = 0.50 mol Ba(OH) 2 x 1 mol BaBr 2 = 0.5 mol BaBr 2 1 mol mol Ba(OH) 2 Therefore, Ba(OH) 2 is the limiting reagent. Find the limiting reagent. a. HBr c. BaBr 2 b. Ba(OH) 2 d. HBr and Ba(OH) In a reaction of 15.0 g of silver nitrate with a large excess of sodium carbonate, the student collected 3.00 g of silver carbonate. The percent yield is.percent YIELD 2 AgNO 3 + Na 2 CO 3 Ag 2 CO NaNO g/mol g/mol g/mol g/mol? g Ag 2 CO 3 theoretical yield = 15.0 g AgNO 3 x 1 mol AgNO 3 x 1 mol Ag 2 CO 3 x g Ag 2 CO 3 = g AgNO 3 2 mol AgNO 3 1 mol Ag 2 CO 3 = 12.2 AgNO 3 so 3.00/ *100 = 24.6% a % c % b % d % 4

5 Bonus Points. All or nothing (2 pts each). AgNO 3 + NaCl AgCl + NaNO g/mol g/mol g/mol g/mol Given: 25.0 grams of AgNO grams of NaCl 1) Determine the mass of NaNO 3 (theoretical yield).? g NaNO 3 = 25.0 g AgNO 3 x 1 mol AgNO 3 x 1 mol NaNO 3 x g NaNO 3 = 12.5 g NaNO g AgNO 3 1 mol AgNO 3 1 mol NaNO 3? g NaNO 3 = 25.0 g NaCl x 1 mol NaCl x 1 mol NaNO 3 x g NaNO 3 = 36.4 g NaNO g NaCl 1 mol NaCl 1 mol NaNO 3 AgNO 3 is the limiting reagent. Hence amount of NaNO 3 produced is 12.5 g. 2) Determine the mass of NaNO 3 produced if the actual yield is 50.0%.? g NaNO 3 actual yield = g NaNO 3 x 0.50 = g NaNO 3 or 6.25 g NaNO 3 3) Calculate the grams of NaCl left after the reaction (based on the 50.0% actual yield of production)? g NaCl used in the reaction = g NaNO 3 x 1 mol NaNO 3 x 1 mol NaCl x g NaCl g NaNO 3 1 mol NaNO 3 1 mol NaCl? g NaCl used in the reaction = g NaCl? g NaCl left after the reaction = initial g NaCl g NaCl used in the reaction = 25.0 g NaCl g NaCl = 20.7 g NaCl Amount of excess NaCl is 20.7 g. 5