Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations


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1 CHE11 Chapter Chapter Stoichiometry: Calculations with Chemical Formulas and Equations 1. When the following equation is balanced, the coefficients are. NH (g) + O (g) NO (g) + H O (g) (a). 1, 1, 1, 1 (b).,,, (c)., 7,, 6 (d). 1,, 1, Explanation: When dealing with any equation with molecular O avoid using fractional coefficients.. When the following equation is balanced, the coefficients are. (a)., 6,, (b)., 1,, (c). 1, 1, 1, 1 (d).,, 1, 6 Al(NO ) + Na S Al S + NaNO. When the following equation is balanced, the coefficient of Al O is. Al O (s) + C (s) + Cl (g) AlCl (s) + CO (g) (a) (b). (c). (d). 1. When the following equation is balanced, the coefficient of H S is. FeCl (aq) + H S (g) Fe S (s) + HCl (aq) (a). 1 (b). (c). (d). 5 Copyright 006 Dr. Harshavardhan D. Bapat 1
2 CHE11 Chapter 5. When the following equation is balanced, the coefficient of HCl is. (a). 0.5 (b). (c). (d). CaCO (s) + HCl (aq) CaCl (aq) + CO (g) + H O (1) 6. When the following equation is balanced, the coefficient of C H 8 O is. C H 8 O (g) + O (g) CO (g) + H O (g) (a). (b). (c). (d) When the following equation is balanced, the coefficient of O is. C H O (g) + O (g) CO (g) + H O (g) (a). 5 (b). (c). (d). 8. When the following equation is balanced, the coefficient of hydrogen is. K (s) + H O (l) KOH (aq) + H (g) (a). (b). 1 (c). (d). 9. When the following equation is balanced, the coefficient of dinitrogen pentoxide is. N O 5 (g) + H O (l) HNO (aq) (a). (b). (c). (d). 1 Copyright 006 Dr. Harshavardhan D. Bapat
3 CHE11 Chapter 10. When the following equation is balanced, the coefficient of nitric acid is. (a). 5 (b). (c). (d). N O 5 (g) + H O (l) HNO (aq) 11. The balanced equation for the decomposition of sodium azide is. (a). NaN (s) (b). NaN (s) (c). NaN (s) (d). None of the above Na (s) + N (g) Na (s) + N (g) Na (s) + N (g) 1. How many moles of carbon atoms are in mol of dimethylsulfoxide (C H 6 SO)? (a). (b). 6 (c). 8 (d). Explanation: This is based on reading the formula and correctly extracting information from it. The formula C H 6 SO indicates that every mole of this compound has moles of carbon atoms in it. Thus moles of the compound would have x = 8 moles of C atoms. 1. There are sulfur atoms in 5 molecules of C H S. (a). 1.5 x 10 5 (b)..8 x 10 5 (c)..0 x 10 (d). 50 Explanation: The molecular formula indicates that every molecule of C H S has sulfur atoms per molecule and hence 5 molecules of this compound will have 5 x = 50 atoms of sulfur. 1. There are hydrogen atoms in 5 molecules of C H S. (a). 5 (b)..8 x 10 (c). 6.0 x 10 5 (d). 100 Copyright 006 Dr. Harshavardhan D. Bapat
4 CHE11 Chapter Explanation: The formula of C H S indicates that there are hydrogen atoms per molecule and hence 100 hydrogen atoms in 5 molecules of C H S. 15. How many carbon atoms are contained in a sample of C H 8 O that contains 00 molecules? (a). 600 (b). 00 (c)..61 x 10 6 (d). 1.0 x How many grams of oxygen are in 65.0 g of C H O? (a). 18 (b). 9 (c). 9.5 (d). 5.8 Explanation: This question uses the mole to mole ratio between oxygen and C H O and needs the following steps g C H O moles O $ $ g O = 5.8 g of O 58.0 g" mol #1 1 mole C H O 1 mole of O 17. How many moles of carbon dioxide are there in 5.06 g of carbon dioxide? (a) (b) (c) x 10 (d) x 10 Explanation: This is a straightforward conversion from grams to moles of CO which is done as follows: 1mole CO 5.06 g CO = 1.18 moles of CO.99 g CO 18. How many moles of the compound magnesium nitrate, Mg(NO ), are in a.5 g sample of this compound? (a) 8. (b) 65.8 (c) (d) Explanation: This is a straightforward conversion from grams to moles of Mg(NO ) which is done as follows: Copyright 006 Dr. Harshavardhan D. Bapat
5 CHE11 Chapter 1mole Mg(NO ).5 g Mg(NO ) = moles g 19. A 5.5g sample of ammonium carbonate contains mol of ammonium ions. (a) (b) (c) (d)..00 Explanation: Realize that the formula for ammonium carbonate is (NH ) CO and calculate the molar mass ( g/mol). Convert grams to moles and then using the stoichiometric ratio find the # of moles of ammonium ions. + 1mol( NH ) CO moles NH 5.5 g ( NH ) CO = 0.51 moles g 1mol NH CO ( ) 0. Magnesium and nitrogen react in a combination reaction to produce magnesium nitride: Mg + N Mg N In a particular experiment, a 5.7g sample of N reacts completely. How many grams of Mg are needed for this reaction? (a). 1. g (b)..1 g (c) g (d). 0.9 g Explanation: Ensure that the equation is balanced. The grams of N must be converted to moles of N and then using the stoichiometric ratio between the Mg and N, the grams of Mg can be calculated. 5.7 g N 1mole N mole Mg.050 g Mg 8.01 g 1mole N 1mole Mg = 1. g Mg 1. The combustion of ammonia in the presence of excess oxygen yields NO and H O: NH (g) + 7O (g) NO (g) + 6H O (g) How many grams of oxygen will the combustion of 08.5 g of ammonia consume? (a). 9.9 g (b). 5.1 g (c)..9 g Copyright 006 Dr. Harshavardhan D. Bapat 5
6 CHE11 Chapter (d) g Explanation: Ensure that the equation is balanced and then convert the grams of ammonia to moles. Using the stoichiometric ratio between ammonia and oxygen calculate the grams of oxygen required. Pay attention to the molar masses being use here. 1mol NH 7 moles O g O 08.5 g NH = g O 17.01g moles NH 1mole O. The combustion of ammonia in the presence of excess oxygen yields NO and H O: NH (g) + 7O (g) NO (g) + 6H O (g) How many grams of NO will be produced by the combustion of 08.5 g of ammonia? (a) g (b) g (c). 56. g (d). 95. g Explanation: Ensure that the equation is balanced and then convert the grams of ammonia to moles. Using the stoichiometric ratio between ammonia and NO calculate the grams of NO required 1mol NH moles NO g NO 08.5 g NH = 56. g NO 17.01g moles NH 1mole NO. Calcium carbide (CaC ) reacts with water to produce acetylene (C H ): CaC (s) + H O (g) Ca(OH) (s) + C H (g) How many g of H O will be required for the production of.5 kg of C H? (a). 500 g (b). 9.0 g (c) g (d).8 x 10 g Explanation: Ensure that the equation is balanced. It is important to convert the kilograms of C H to grams first and then to moles of acetylene. Using the stoichiometric factor between acetylene and water the grams of water can be calculated. 1mole moles H O g.5 10 g C H =.8 10 g H O 6.08 g 1mole C H 1mole Copyright 006 Dr. Harshavardhan D. Bapat 6
7 CHE11 Chapter. Under appropriate conditions, nitrogen and hydrogen react to yield ammonia: N (g) + H (g) NH (g) How many grams of hydrogen will be needed to react completely with 75.8 g of N? (a). 51. g (b) 16. g (c). 11 g (d). 5. g Explanation: Ensure that the equation is balanced. Convert the grams of nitrogen to moles of nitrogen. Using the stoichiometric factor between nitrogen and hydrogen the grams of hydrogen can be calculated. 1mole N moles H.016 g H 75.8 g N = 16. g H 8.01 g 1mole N 1mole H 5. Lead (II) carbonate decomposes to give lead (II) oxide and carbon dioxide: PbCO (s) PbO (s) + CO (g) How many grams of lead (II) carbonate should decompose completely to produce 0.0 g of lead (II) oxide? (a). 0.1 g (b). 0.0 g (c).9 g (d). 0.9 g Explanation: Ensure that the equation is balanced. Convert the grams of lead (II) oxide to moles of lead (II) oxide. Using the stoichiometric factor between lead (II) oxide and lead (II) carbonate the grams of lead (II) carbonate can be calculated. 1mole PbO 1mole PbCO g PbCO 0.0 g PbO =.9 g PbCO.199 g 1mole PbO 1mole PbCO 6. GeF H is formed from GeH and GeF in the combination reaction: GeH + GeF GeF H If the reaction yield is 9.6%, how many moles of GeF are needed to produce 8.00 mol of GeF H? (a).. (b). 6.8 (c) Copyright 006 Dr. Harshavardhan D. Bapat 7
8 CHE11 Chapter (d)..16 Explanation: This question is based on the ideas of actual, theoretical and percent yields. According to the balanced equation moles of GeF would produce moles of GeF H if the yield was 100% (which would be the theoretical yield). The % yield is only 9.6% meaning only 9.6% of moles of GeF H (0.96 x =.70 moles) will be produced. To produce 8 moles of GeF H, theoretically 6 moles of GeF will be required, but since the %yield is only 9.6%, the # of moles of GeF required = 8.00 moles GeF H moles GeF = 6.8 moles GeF.70 moles GeF H 7. When a substance burns in air, what component of air reacts? (a). oxygen (b). nitrogen (c). carbon dioxide (d). water 8. Of the reactions below, which one is not a combination reaction? (a). C + O CO (b). Mg + O MgO (c). CaO + H O Ca (OH) (d). CH + O CO + H O 9. Of the reactions below, which one is a decomposition reaction? (a). NH Cl NH + HCl (b). Mg + O MgO (c). N + H NH (d). Cd(NO ) + Na S CdS + NaNO 0. Which of the following are combination reactions? (1). CH (g) + O (g) CO (g) + H O (l) (). CaO (s) + CO (g) CaCO (s) (). Fe (s) + O (g) Fe O (s) (). PbCO (s) PbO (s) + CO (g) (a). 1,, and only (b). and only (c). All of them (d).,, and only Copyright 006 Dr. Harshavardhan D. Bapat 8
9 CHE11 Chapter 1. The mass % of Al in aluminum sulfate (Al (SO ) ) is. (a) (b). 1.9 (c) (d) Explanation: Calculate the formula mass of aluminum sulfate first (=.1 g/mol) and then divide the mass of the aluminum present in aluminum sulfate by the formula mass. Don t forget to multiply by 100% Molar mass of Al (6.98 g/mol) 100% = 15.77% Formula mass of aluminum sulfate (.1 g/mol). The formula weight of a substance is. (a). the same as the percent by mass weight (b). determined by combustion analysis (c). the sum of atomic weights of each atom in its chemical formula (d). the weight of a sample of the substance.. The mass % of C in methane (CH ) is. (a). 5.1 (b). 1.6 (c). 9.6 (d) Explanation: Calculate the formula mass of methane first (= 16.0 g/mol) and then divide the mass of the carbon present in the methane by the formula mass. Don t forget to multiply by 100% ' 1.01g/mol (mass of C present) $ % " 100% = 7.87% & 16.0g/mol (molar mass of methane) #. One mole of contains the largest number of atoms. (a). S 8 (b). C 10 H 8 (c). Al (SO ) (d). Na PO Explanation: This question is based on the definition of a mole and the number of atoms in a given formula. One mole of molecules = 6.0 x 10 molecules. Since the molecules of each substance here are made of a different number of atoms, the one with the largest number of atoms in its formula (18 in C 10 H 8 ) will be the answer. Copyright 006 Dr. Harshavardhan D. Bapat 9
10 CHE11 Chapter 5. One million argon atoms is mol of argon atoms. (a). 1.7 x (b). 6.0 x 10 (c). 1.7 x (d). 1.0 x Explanation: By definition, one mole of argon atoms = 6.0 x 10 atoms. The number of moles of argon atoms equal to 1 million argon atoms can then be calculated by: 6 1.0" 10 atoms of Ar 18 = 1.7 " 10 moles of Argon atoms 6.0" 10 atoms/mole 6. There are atoms of oxygen in 00 molecules of CH CO H. (a). 600 (b). 150 (c)..01 x 10 (d)..61 x 10 6 Explanation: Since there are atoms of oxygen per molecule of CH CO H, there are x 00 = 600 atoms of oxygen present. 7. How many molecules of CH are in 8. g of this compound? (a) x 10 (b)..00 (c)..90 x 10 5 (d) x 10 Explanation: Convert the grams of methane to moles of methane first and then using the definition of a mole, calculate the # of molecules. Since the question is asking for the number of molecules in 8. g of methane, the answer will be a large number. 1mole CH molecules 8. g CH = molecules of CH 16.0 g 1mole CH 8. A sample of CH F with a mass of 0.0 g contains atoms of F. (a).. x 10 (b). 0.0 (c)..6 x 10 (d)..6 x 10  Copyright 006 Dr. Harshavardhan D. Bapat 10
11 CHE11 Chapter Explanation: Convert the grams of CH F to moles first and then using the definition of a mole, calculate the # of molecules. Now use the number of atoms of F per mole to find the number of fluorine atoms. Since the question is asking for the number of atoms in 0.0 g of CH F, the answer will be a large number. 1mole CHF molecules CHF atoms F 0.0 g CHF 5.0 g CH F 1mole CH F 1 molecule CH F =.6 10 atoms F 9. How many atoms of nitrogen are in g of NH NO? (a)..5 (b) x 10 (c) x 10 (d). 1.8 Explanation: Convert the grams of NH NO to moles first and then using the definition of a mole, calculate the # of molecules. Now use the number of atoms of N per mole to find the number of nitrogen atoms. Since the question is asking for the number of atoms of N in g of NH NO, the answer will be a large number. 1mole NHNO molecules of NHNO atoms g NHNO 80.0 g NH NO 1mole NH NO 1mole NH = atoms of N 0. What is the mass in grams of 9.76 x atoms of magnesium? (a)..0 (b). 1.6 x (c)..9 x 10 (d). None of the above Explanation: Convert the number of atoms to number of moles and then using the molar mass of magnesium, calculate the number of grams. As the number of atoms is being converted to grams the answer will be a small number. 18 1mole Mg atoms.05 g Mg 9.76" 10 atoms Mg " " =.9" 10 g of Mg 6.0" 10 atoms 1mole Mg N NO Copyright 006 Dr. Harshavardhan D. Bapat 11
12 CHE11 Chapter 1. Calculate the number of sulfur dioxide molecules in 1.58 moles of sulfur dioxide. How many molecules of oxygen is this number equal to? (a) x 10 (b). 6.0 x 10  (c) x 10 (d) x 10 Explanation: Convert the number of moles to number of molecules of SO and then to atoms of oxygen. Need to find the number of molecules of oxygen after this molecules atoms of O 1molecule O 1.58 moles of SO 1mole of SO 1 molecule of SO atoms of O = molecules of O. A nitrogen oxide is 0.% by mass nitrogen. The molecular formula could be. (a). NO (b). N O (c). either NO or N O (d). NO Explanation: Calculate the mass percent of nitrogen in each of the oxides based on the molar mass of the oxide and the mass of nitrogen present in it For NO, mass % of N = 100% = 6.7% Doing similar calculations for the rest N O : 0.%, NO : 0.%. A sulfur oxide is 50.0% by mass sulfur. It s molecular formula could be. (a). SO (b). SO (c). S (d). Both SO or S O Explanation: Calculate the mass percent of sulfur in each of the oxides based on the molar mass of the oxide and the mass of sulfur present in it. SO is one of the correct answers but so is S O, so do not stop at just the SO..065 For SO,mass % of S = 100% = 50.0% Doing similar calculations for the others SO = 66.7% and S O = 50.0%. Copyright 006 Dr. Harshavardhan D. Bapat 1
13 CHE11 Chapter. Which hydrocarbon pair below has identical mass percentage of C? (a). C H and C H 6 (b). C H and C H (c). C H and C H 6 (d). C H and C H Explanation: Calculate the mass percent of carbon in each of the hydrocarbons based on the molar mass of the hydrocarbon and the mass of carbon present in it..0 For C H,mass % of C = 100% = 85.6% 8.05 Doing similar calculations for others, C H 6 = 85.6%, C H = 89.9% and C H = 95.9% Copyright 006 Dr. Harshavardhan D. Bapat 1
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