Stoichiometry. Lecture Examples Answer Key


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1 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 N H 2 O + 9 O 2 2 Ca(OH) SO 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 = g mol 1 (b) Ca(C 2 H 3 O 2 ) 2 MW = g mol 1 (c) [K 2 (UO 2 ) 2 (VO 4 ) 2 3H 2 O] MW = g mol 1 Ex. 5 Calculate the percentage by mass of oxygen in 4a and 4b above. (a) [( g mol 1 ) / g mol 1 ] 100% = % (b) [( g mol 1 ) / g mol 1 ] 100% = % Ex. 6 (a) Calculate the number of O atoms in mol of C 6 H 12 O 6. # O atoms = (0.470 mol C 6 H 12 O 6 )( molecules/1 mol)(6 O atoms / 1 molecule) = O atoms (b) Calculate the total number of ions in 38.1 g of CaF 2. # ions = (38.1 g CaF 2 )(1 mol / g)( molecules/1 mol) (3 ions / 1 molecule) = ions Ex. 7 (a) What is the mass in grams of 1 mol of glucose, C 6 H 12 O 6? mass = (1 mol)( g/1 mol) = g (b) What is the mass in grams of 3.52 mol of chromium(iii) sulfate decahydrate? 1
2 [Cr 2 (SO 4 ) 3 10H 2 O] mass = (3.52 mol)( g/1 mol) = 2015 g Ex. 8 How many moles of chloride ions are in g of magnesium chloride? moles Cl = ( g MgCl 2 )(1 mol MgCl 2 / g)(2 mol Cl /1 mol MgCl 2 ) = mol Cl Ex. 9 What is the mass, in grams, of molecules of caffeine, C 8 H 10 N 4 O 2? mass = ( molecules)(1 mol/ molecules)( g/1 mol) = g Ex. 10 What is the molar mass of cholesterol if mol weighs g? Molar mass = g / mol = 387 g mol 1 Ex. 11 Determine the empirical formula of a compound containing mol K, mol C, and mol O. K ==> mol / mol = 3 C ==> mol / mol = 1 K 3 CO 3 O ==> mol / 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 = ( g CO 2 )(1 mol CO 2 / g)(1 mol C/1 mol CO 2 ) = mol C moles H = ( g H 2 O)(1 mol H 2 O/ g)(2 mol H/1 mol H 2 O) = mol H Since toluene contains only C and H (we are told in the problem) we have everything we need. C ==> mol / mol = H ==> mol / mol = C 7 H 8 2
3 Ex. 14 A sample of 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  )( 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 O 2 12 CO H 2 O mass O 2 = (1.00 g C 3 H 9 )(1 mol C 3 H 9 / g)(21 mol O 2 /4 mol C 3 H 9 )( 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/ g) = mol H 2 S moles NaOH = (1.85 g NaOH)(1 mol NaOH/ g) = mol NaOH moles of Na 2 S formed if H 2 S is limiting = ( mol H 2 S)(1 mol Na 2 S / 1 mol H 2 S) = mol Na 2 S moles of Na 2 S formed if NaOH is limiting = ( mol NaOH)(1 mol Na 2 S / 2 mol NaOH) = mol Na 2 S 3
4 The smaller amount of Na 2 S formed will tell us the limiting reactant. NaOH is limiting reactant. mass of Na 2 S = ( mol Na 2 S)( 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 / g) = mol C 2 H 6 moles Cl 2 = (255 g)(1 mol / g) = 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 ) = 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 ) = mol C 2 H 5 Cl Cl 2 is limiting reactant theoretical yield = (3.596 mol Cl 2 H 5 Cl)( 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 / = 86.0 kg moles H 2 O = (86.0 kg)(1 kmol / kg) = kmol H 2 O moles H 2 = (4.774 kmol H 2 O)(2 kmol H 2 / 2 kmol H 2 O) = kmol H 2 mass H 2 = (4.774 kmol H 2 )( 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 lessthanexpected 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 g of each reactant was used. What is the highest percent yield of N 2 that can be expected? 4
5 Here are the numbers of moles of reactants that we start with: moles N 2 O 4 in g = (100.0 g)(1 mol / g) = mol N 2 O 4 moles N 2 H 4 in g = (100.0 g)(1 mol / g) = mol N 2 H 4 Assuming no side reaction occurs: moles N 2 that can be made from N 2 O 4 = ( mol N 2 O 4 )(3 mol N 2 / 1 mol N 2 O 4 ) = mol N 2 possible moles N 2 that can be made from N 2 H 4 = ( mol N 2 H 4 )(3 mol N 2 / 2 mol N 2 H 4 ) = 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) = ( mol N 2 )( 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 / g) = mol NO moles N 2 O 4 used to make 10.0 g NO = ( mol NO)(2 mol N 2 O 4 /6 mol NO) = mol moles N 2 H 4 used to make 10.0 g NO =( mol NO)(1 mol N 2 H 4 /6 mol NO)= 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 = mol mol = mol moles N 2 H 4 available to make N 2 = mol mol = mol moles N 2 that can be made from N 2 O 4 = ( mol N 2 O 4 )(3 mol N 2 / 1 mol N 2 O 4 ) = mol N 2 possible moles N 2 that can be made from N 2 H 4 = ( mol N 2 H 4 )(3 mol N 2 / 2 mol N 2 H 4 ) = 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) = ( mol N 2 )( 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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