Stoichiometry. Lecture Examples Answer Key



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Stoichiometry Lecture Examples Answer Key Ex. 1 Balance the following chemical equations: 3 NaBr + 1 H 3 PO 4 3 HBr + 1 Na 3 PO 4 2 C 3 H 5 N 3 O 9 6 CO 2 + 3 N 2 + 5 H 2 O + 9 O 2 2 Ca(OH) 2 + 2 SO 2 + 1 O 2 2 H 2 O + 2 CaSO 4 Ex. 2 Write a balanced chemical reaction for the combustion of octane, C 8 H 18(l), in air. 2 C 8 H 18(l) + 25 O 2(g) 18 H 2 O (l) + 16 CO 2(g) Ex. 3 Write the balanced chemical equation for the combination of the metallic element calcium with the nonmetallic element oxygen, O 2. 2 Ca (s) + O 2(g) 2 CaO (s) Ex. 4 Calculate the molecular weight of (a) N 2 O 5 MW = 108.0104 g mol 1 (b) Ca(C 2 H 3 O 2 ) 2 MW = 158.167 g mol 1 (c) [K 2 (UO 2 ) 2 (VO 4 ) 2 3H 2 O] MW = 902.1758 g mol 1 Ex. 5 Calculate the percentage by mass of oxygen in 4a and 4b above. (a) [(5 15.9994 g mol 1 ) / 108.0104 g mol 1 ] 100% = 74.0642% (b) [(4 15.9994 g mol 1 ) / 158.167 g mol 1 ] 100% = 40.4620% Ex. 6 (a) Calculate the number of O atoms in 0.470 mol of C 6 H 12 O 6. # O atoms = (0.470 mol C 6 H 12 O 6 )(6.022 10 23 molecules/1 mol)(6 O atoms / 1 molecule) = 1.70 10 24 O atoms (b) Calculate the total number of ions in 38.1 g of CaF 2. # ions = (38.1 g CaF 2 )(1 mol / 78.075 g)( 6.022 10 23 molecules/1 mol) (3 ions / 1 molecule) = 8.82 10 23 ions Ex. 7 (a) What is the mass in grams of 1 mol of glucose, C 6 H 12 O 6? mass = (1 mol)(180.1572 g/1 mol) = 180.1572 g (b) What is the mass in grams of 3.52 mol of chromium(iii) sulfate decahydrate? 1

[Cr 2 (SO 4 ) 3 10H 2 O] mass = (3.52 mol)(572.335 g/1 mol) = 2015 g Ex. 8 How many moles of chloride ions are in 0.0750 g of magnesium chloride? moles Cl = (0.0750 g MgCl 2 )(1 mol MgCl 2 /95.211 g)(2 mol Cl /1 mol MgCl 2 ) = 0.00158 mol Cl Ex. 9 What is the mass, in grams, of 1.75 10 20 molecules of caffeine, C 8 H 10 N 4 O 2? mass = (1.75 10 20 molecules)(1 mol/6.022 10 23 molecules)(194.1926 g/1 mol) = 0.0564 g Ex. 10 What is the molar mass of cholesterol if 0.00105 mol weighs 0.406 g? Molar mass = 0.406 g / 0.00105 mol = 387 g mol 1 Ex. 11 Determine the empirical formula of a compound containing 0.104 mol K, 0.052 mol C, and 0.156 mol O. K ==> 0.104 mol / 0.052 mol = 3 C ==> 0.052 mol / 0.052 mol = 1 K 3 CO 3 O ==> 0.156 mol / 0.052 mol = 3 Ex. 12 What is the molecular formula of a compound that has an empirical formula, CH 2, and a molar mass of 84 g mol 1? Molar mass of CH 2 = 14 g mol 1 84 g mol 1 / 14 g mol 1 = 6 Molecular formula: C 6 H 12 Ex. 13 Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO 2 and 1.37 mg of H 2 O. If the compound contains only carbon and hydrogen, what is its empirical formula? moles C = (5.86 10 3 g CO 2 )(1 mol CO 2 /44.0098 g)(1 mol C/1 mol CO 2 ) = 1.33 10 4 mol C moles H = (1.37 10 3 g H 2 O)(1 mol H 2 O/18.0152 g)(2 mol H/1 mol H 2 O) = 1.52 10 4 mol H Since toluene contains only C and H (we are told in the problem) we have everything we need. C ==> 1.33 10 4 mol / 1.33 10 4 mol = 1 7 7 H ==> 1.52 10 4 mol / 1.33 10 4 mol = 1.144 8 C 7 H 8 2

Ex. 14 A sample of 0.600 mol of a metal M reacts completely with excess fluorine to form 46.8 g of MF 2. (a) How many moles of F - are in the sample of MF 2 that forms? M + F 2 MF 2 mol F = (0.600 mol M)(1 mol MF 2 /1 mol M)(2 mol F - /1 mol MF 2 ) = 1.20 mol F - (b) How many grams of M are in this sample of MF 2? mass of M in this sample will equal the total mass of the sample minus the mass that is due to F -. mass of F - in MF 2 = (1.20 mol F - )(18.9984 g /1 mol F - ) = 22.8 g F - mass of M in MF 2 = 46.8 g 22.8 g F - = 24.0 g M (c) What element is represented by the symbol M? We can determine what M is if we know it s molar mass. We can get that by dividing the mass of M in the sample by the number of moles of M in the sample: molar mass of M = (24.0 g M)(0.600 mol M) = 40.0 g/mol From the periodic table we see that M must be Ca. Ex. 15 Propane, C 3 H 8, is a common fuel used for cooking and home heating. What mass of O 2 is consumed in the combustion of 1.00 g of propane? 4 C 3 H 9 + 21 O 2 12 CO 2 + 18 H 2 O mass O 2 = (1.00 g C 3 H 9 )(1 mol C 3 H 9 /45.1041g)(21 mol O 2 /4 mol C 3 H 9 )(31.9988 g O 2 / 1 mol) = 3.72 g O 2 Ex. 16 g of When hydrogen sulfide gas is bubbled into a solution of sodium hydroxide, the reaction forms sodium sulfide and water. How many grams of sodium sulfide are formed if 2.50 hydrogen sulfide is bubbled into a solution containing 1.85 g of sodium hydroxide, assuming that the limiting reactant is completely consumed? H 2 S + 2 NaOH Na 2 S + 2 H 2 O moles H 2 S = (2.50 g H 2 S)(1 mol H 2 S/34.0818 g) = 0.07335 mol H 2 S moles NaOH = (1.85 g NaOH)(1 mol NaOH/39.9972 g) = 0.04625 mol NaOH moles of Na 2 S formed if H 2 S is limiting = (0.07335 mol H 2 S)(1 mol Na 2 S / 1 mol H 2 S) = 0.07335 mol Na 2 S moles of Na 2 S formed if NaOH is limiting = (0.04625 mol NaOH)(1 mol Na 2 S / 2 mol NaOH) = 0.023125 mol Na 2 S 3

The smaller amount of Na 2 S formed will tell us the limiting reactant. NaOH is limiting reactant. mass of Na 2 S = (0.023125 mol Na 2 S)(78.0458 g/1 mol) = 1.80 g Na 2 S Ex. 17 When ethane, C 2 H 6, reacts with chlorine, Cl 2, the main product is C 2 H 5 Cl, but other products are also obtained in small quantities. The formation of these other products reduces the yield of C 2 H 5 Cl. Calculate the percent yield of C 2 H 5 Cl if the reaction of 125 g of C 2 H 6 with 255 g of Cl 2 produced 206 g of C 2 H 5 Cl. C 2 H 6 + Cl 2 C 2 H 5 Cl + HCl First calculate the theoretical yield: moles C 2 H 6 = (125 g)(1 mol / 30.0694 g) = 4.157 mol C 2 H 6 moles Cl 2 = (255 g)(1 mol / 70.906 g) = 3.596 mol Cl 2 moles C 2 H 5 Cl if C 2 H 6 is limiting = (4.157 mol C 2 H 6 )(1 mol C 2 H 5 Cl/ 1 mol C 2 H 6 ) = 4.157 mol C 2 H 5 Cl moles C 2 H 5 Cl if Cl 2 is limiting = (3.596 mol Cl 2 )(1 mol C 2 H 5 Cl/ 1 mol Cl 2 ) = 3.596 mol C 2 H 5 Cl Cl 2 is limiting reactant theoretical yield = (3.596 mol Cl 2 H 5 Cl)(64.5145 g / 1 mol) = 232 g C 2 H 5 Cl % yield = (actual yield / theoretical yield) 100% = (206 g/232 g) 100% = 88.8% Ex. 18 Hydrogen gas has been suggested as a clean fuel because it produces only water vapor when it burns. If the reaction has a 98.8% yield, what mass of hydrogen forms 85.0 kg of water? 2 H 2 + O 2 2 H 2 O The actual yield is 85.0 kg. Using this and the percentage yield we can find the theoretical yield: theoretical yield = 85.0 kg / 0.988 = 86.0 kg moles H 2 O = (86.0 kg)(1 kmol / 18.0152 kg) = 4.774 kmol H 2 O moles H 2 = (4.774 kmol H 2 O)(2 kmol H 2 / 2 kmol H 2 O) = 4.774 kmol H 2 mass H 2 = (4.774 kmol H 2 )(2.0158 kg H 2 / 1 kmol H 2 ) = 9.62 kg H 2 Ex. 19 During studies of the reaction N 2 O 4 (l) + 2 N 2 H 4 (l) 3 N 2 (g) + 4 H 2 O(g) a chemical engineer measured a less-than-expected yield of N 2 and discovered that the following side reaction occurs: 2 N 2 O 4 (l) + N 2 H 4 (l) 6 NO(g) + 2 H 2 O(g). In one experiment, 10.0 g of NO formed when 100.0 g of each reactant was used. What is the highest percent yield of N 2 that can be expected? 4

Here are the numbers of moles of reactants that we start with: moles N 2 O 4 in 100.0 g = (100.0 g)(1 mol / 92.011 g) = 1.0868 mol N 2 O 4 moles N 2 H 4 in 100.0 g = (100.0 g)(1 mol / 32.045 g) = 3.1206 mol N 2 H 4 Assuming no side reaction occurs: moles N 2 that can be made from N 2 O 4 = (1.0868 mol N 2 O 4 )(3 mol N 2 / 1 mol N 2 O 4 ) = 3.2604 mol N 2 possible moles N 2 that can be made from N 2 H 4 = (3.1206 mol N 2 H 4 )(3 mol N 2 / 2 mol N 2 H 4 ) = 4.6809 mol N 2 possible N 2 O 4 is the limiting reactant This gives us the theoretical yield of N 2 (if side reaction does not occur) = (3.2604 mol N 2 )(28.0134 g / 1 mol) = 91.3 g N 2 Since we know the side reaction does occur we need to know how much of each reactant is used up in the reaction (because that means we have that much less reactant available to make N 2 ): moles NO in 10.0 g = (10.0 g NO)(1 mol / 30.0061 g) = 0.33327 mol NO moles N 2 O 4 used to make 10.0 g NO = (0.33327 mol NO)(2 mol N 2 O 4 /6 mol NO) =0.1111 mol moles N 2 H 4 used to make 10.0 g NO =(0.33327 mol NO)(1 mol N 2 H 4 /6 mol NO)=0.05554 mol Now we can calculate the number of moles of each reactant left over to make N 2 : moles N 2 O 4 available to make N 2 = 1.0868 mol 0.1111 mol = 0.9757 mol moles N 2 H 4 available to make N 2 = 3.1206 mol 0.05554 mol = 3.0651 mol moles N 2 that can be made from N 2 O 4 = (0.9757 mol N 2 O 4 )(3 mol N 2 / 1 mol N 2 O 4 ) = 2.9271 mol N 2 possible moles N 2 that can be made from N 2 H 4 = (3.0651 mol N 2 H 4 )(3 mol N 2 / 2 mol N 2 H 4 ) = 4.5976 mol N 2 possible N 2 O 4 is the limiting reactant (still) Maximum yield of N 2 possible, since side reaction does occur (we will use this as our actual yield) = (2.9271 mol N 2 )(28.0134 g / 1 mol) = 82.0 g N 2 Maximum percentage yield = (82.0 g / 91.3 g) 100% = 89.9% 5

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