= mol H O mol H O. 10 mol H O x g C H. 10 mol H O x g O. = 75 mol O. = 34.5 mol O
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1 1. C 4 H O þ 8CO + 10H O.46 g a).46 g H O x 1 mol H O 18.0 g H O = mol H O b).46 g H O x 1 mol H O 18.0 g H O x mol C H mol H O = mol C H 4 10 c).46 g H O x 1 mol H O 18.0 g H O x mol C H 10 mol H O x g C H mol C H = 1.59 g C H 4 10 d).46 g H O x 1 mol H O 18.0 g H O x 13 mol O 10 mol H O x 3.00 g O 1 mol O = 5.68 g O. C H 6 O + 3O þ CO + 3H O a) 5 mol C H O x 6 3 mol O 1 mol C H O 6 = 75 mol O b) 30 mol O x 1 mol C H O 6 3 mol O = 10 mol C H O 6 30 mol O x mol CO 3 mol O = 0 mol CO c) 3 mol CO x 3 mol O mol CO = 34.5 mol O d) 41 mol H O x 3 mol O 3 mol H O = 41 mol O 41 mol H O x mol CO 3 mol H O = 7.3 mol CO
2 3. Fe O 3 (s) + 3H (g) þ Fe(s) + 3H O(l) a) 5 mol Fe O x 3 mol Fe 1 mol Fe O 3 = 50 mol Fe b) 30 mol Fe x 3 mol H mol Fe = 45 mol H c) 10 mol H O x 1 mol Fe O 3 mol H O g Fe O x 1 mol Fe O 3 3 = 6388 g Fe O 3 4. N H 4 + 7H O þ HNO 3 + 8H O a).68 mol N H x 7 mol H O 4 1 mol N H 4 = mol H O b).68 mol N H x 4 mol HNO 1 mol N H 4 3 = 5.36 mol HNO 3 c).68 mol N H x 8 mol H O 4 1 mol N H 4 = 1.44 mol H O 5. WO 3 (s) + 3H (g) ==> W(s) + 3H O(l) 50 g? g? g a) 50 g WO x 3 1 mol WO g WO x 1 mol W 1 mol WO x g W 3 1 mol W = 198 g W 3 3 b) 50 g WO x 3 1 mol WO g WO x 3 mol H 1 mol WO x.0 g H 3 1 mol H = 6.53 g H 3 3
3 STOICHIOMETRIC PROBLEMS #3 - ANSWERS 1. C 4 H O þ 8 CO + 10 H O? g 100 g 100 g H O x 1 mol H O x mol C 4 H 10 x g C 4 H 10 = g C 4 H g H O 10 mol H O 1 mol C 4 H 10. KClO 3 þ KCl + 3 O.50 mol? mol.50 mol KClO 3 x 3 mol O = 3.75 mol O mol KClO Ca(NO 3 ) + K 3 PO 4 þ Ca 3 (PO 4 ) + 6 KNO g? g? g? g g Ca(NO 3 ) x 1 mol Ca(NO 3 ) x 1 mol Ca 3 (PO 4 ) x g Ca 3 (PO 4 ) = g Ca 3 (PO 4 ) g Ca(NO 3 ) 3 mol Ca(NO 3 ) 1 mol Ca 3 (PO 4 ) g Ca(NO 3 ) x 1 mol Ca(NO 3 ) x mol K 3 PO 4 x 1.7 g K 3 PO 4 = g K 3 PO g Ca(NO 3 ) 3 mol Ca(NO 3 ) 1 mol K 3 PO g Ca(NO 3 ) x 1 mol Ca(NO 3 ) x 6 mol KNO 3 x g KNO 3 = g KNO g Ca(NO 3 ) 3 mol Ca(NO 3 ) 1 mol KNO 3
4 4. mass reactants = mass Ca(NO 3 ) + mass K 3 PO 4 = g g = g mass products = mass Ca 3 (PO 4 ) + mass KNO 3 = g g = g The mass of reactants and products are the same (or nearly the same). This makes sense, since the law of conservation of mass states that matter cannot be created nor destroyed, therefore mass is conserved. 5. Al O H O þ Al(OH) g? g 5.0 g Al O 3 x 1 mol Al O 3 x mol Al(OH) 3 x g Al(OH) 3 = 38.3 g Al(OH) g Al O 3 1 mol Al O 3 1 mol Al(OH) 3 6. CuO þ Cu O + O 35.0 g? g 35.0 g CuO x 1 mol CuO x 1 mol Cu O x g Cu O = 31.5 g Cu O g CuO mol CuO 1 mol Cu O 7. Ca(ClO 3 ) þ CaCl + 3 O 11.0 mol? g? mol 11.0 mol Ca(ClO 3 ) x 3 mol O = 33.0 mol O 1 mol Ca(ClO 3 ) 11.0 mol Ca(ClO 3 ) x 1 mol CaCl x g CaCl = 10 g CaCl 1 mol Ca(ClO 3 ) 1 mol CaCl
5 8. Cu + AgNO 3 þ Cu(NO 3 ) + Ag? mol 1.5 mol? g 13 g 1.5 mol AgNO 3 x 1 mol Cu = 0.65 mol Cu mol AgNO 3 13 g Ag x 1 mol Ag x 1 mol AgNO 3 x g AgNO 3 = 08 g AgNO g Ag 1 mol Ag 1 mol AgNO 3 9. Fe O C þ Fe + 3 CO 5.0 kg? kg 5.0 kg Fe O 3 x 1000 g x 1 mol Fe O 3 x mol Fe 1 kg g Fe 1 mol Fe O 3 x g Fe x 1 kg = 17.5 kg Fe 1 mol Fe 1000 g 10. C 6 H 1 O O þ 6 CO + 6 H O 10 g? g 10 g C 6 H 1 O 6 x 1 mol C 6 H 1 O 6 x 6 mol CO x g CO = 176 g CO g C 6 H 1 O 6 1 mol C 6 H 1 O 6 1 mol CO The mass of oxygen added to carbon in carbon dioxide gives this product more mass when compared to the mass of oxygen (and hydrogen) added to the carbon in glucose.
6 11. 4 NH O þ 4 NO + 6 H O 14.0 g? mol? g 14.0 g NH 3 x 1 mol NH 3 x 6 mol H O x 18.0 g H O =. g H O g NH 3 4 mol NH 3 1 mol H O 14.0 g NH 3 x 1 mol NH 3 x 5 mol O = 1.03 mol O g NH 3 4 mol NH 3
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13 SHEET 6 ANSWERS g of Ca C reacts with 4.0 g of H O according to the following reaction: CaC (s)+h O(l) C H (g)+ca(oh) (s) a) Determine which reactant in the limiting reagent. b) What mass of C H (g) and Ca(OH) (s) is produced. c) Calculate the excess mass of the excess reagent CaC (s) 16.0 g + H O(l) 4.0 g C H (g)? g + Ca(OH) (s)? g a) This question is usually not asked directly. If there is information given about two or more reactants, this step MUST BE TAKEN. Consider CaC : Consider H O : 16.0 g CaC x 1 mol CaC g CaC = 0.50 mol CaC available 0.50 mol CaC x mol H O 1 mol CaC = mol H O required 4.0 g H O x 1 mol H O 18.0 g H O =.33 mol H O available.3 mol H O x 1 mol CaC mol H O = 1.17 mol CaC required therefore the limiting reagent is CaC b) These are examples of typical final questions mol CaC x 1 mol C H 1 mol CaC x 6.04 g C H 1 mol C H = 6.50 g C H produced 0.50 mol CaC x 1 mol Ca(OH) 1 mol CaC x g Ca(OH) 1 mol Ca(OH) = 18.5 g Ca(OH) produced c) This is not a typical question but it helps to point out that there will be left overs for the excess reagent mol CaC x mol H O 1 mol CaC x 18.0 g H O 1 mol H O = 9.00 g H O consumed mass H O excess = (mass H O available) (mass H O consumed) = (4.0 g H O) (9.00 g H O) = 33.0 g H O remains after reaction
14 . Consider the following reaction at S.T.P. If 35 g of tungsten trioxide reacts with 15 L of H at S.T.P., what mass of tungsten is produced? WO 3 (s) 35.0 g + 3H (g) 15.0 L W(s)? g + 3H O(aq) Consider WO 3 : Consider H : 35.0 g WO 3 x 1 mol WO g WO 3 = mol WO 3 available mol WO 3 x 3 mol H 1 mol WO 3 = mol H required 15.0 L H x 1 mol H.414 L H = mol H available mol H x 1 mol WO 3 3 mol H = 0.3 mol WO 3 required therefore the limiting reagent is WO mol WO 3 x 1 mol W 1 mol WO 3 x g W 1 mol W = 7.8 g W
15 3. What mass of H SO 4 can be produced from 50.0 g of SO, 15.0 g O and an unlimited amount of H O? The equation is: Consider SO : Consider O : SO (g) 50.0 g + O (g) 15.0 g + H O(l) H SO 4 (aq)? g 50.0 g SO x 1 mol SO g SO = mol SO available mol SO x 1 mol O mol SO = mol O required 15.0 g O x 1 mol O 3.00 g O = mol O available mol O x mol SO 1 mol O = mol SO required therefore the limiting reagent is SO mol SO x mol H SO 4 mol SO x g H SO 4 1 mol H SO 4 = 76.5 g H SO 4
16 L of O react with 19.6 L of methane (CH 4 ) at S.T.P. according to the reaction shown below. What volume of water and carbon dioxide are produced at S.T.P. CH 4 (g) 19.6 L + O (g) 40.0 L CO (g) + H O(g)? S.T.P.? S.T.P. Consider CH 4 : Consider O : 19.6 L CH 4 x 1 mol CH L CH 4 = mol CH 4 available mol CH 4 x mol O 1 mol CH 4 = 1.75 mol O required 40.0 L O x 1 mol O.414 L O = 1.78 mol O available 1.78 mol O x 1 mol CH 4 mol O = 0.89 mol CH 4 required therefore the limiting reagent is CH mol CH 4 x mol H O 1 mol CH 4 x.414 L H O 1 mol H O = 39. L H O mol CH 4 x 1 mol CO 1 mol CH 4 x.414 L CO 1 mol CO = 19.6 L CO
17 5. What is the maximum mass of carbon dioxide that can be produced by the reaction between 15.0 g of propane (C 3 H 8 ) with 60.0 g of oxygen gas? C 3 H 8 (g) 15.0 g + 5O (g) 60.0 g 3CO (g)? g 3CO (g) + 4H O(g) Consider C 3 H 8 : Consider O : 15.0 g C 3 H 8 x 1 mol C 3H g C 3 H 8 = mol C 3 H 8 available mol C 3 H 8 x 5 mol O 1 mol C 3 H 8 = 1.70 mol O required 60.0 g O x 1 mol O 3.00 g O = 1.88 mol O available 1.88 mol O x 1 mol C 3 H 8 5 mol O = mol C 3 H 8 required therefore the limiting reagent is C 3 H mol C 3 H 8 x 3 mol CO 1 mol C 3 H 8 x g CO 1 mol CO = 44.9 g CO
18 6. What mass of iron(iii) oxide is produced when 0.9 g of iron(ii) sulphide reacts with 10.0 L of oxygen gas at kpa and a temperature of 4 C? What volume of sulphur dioxide is produced at S.T.P.? Consider O : 4FeS(s) 0.9 g P = kpa V = 10.0 L n =? R = kpa L K mol T = 4 o C K + 7O (g) 10.0 L kpa 4 o C Fe O 3 (s)? g + 4SO (g)? S.T.P. n = PV RT n = kpa 10.0 L kpa L K K mol n = mol O available mol O x 4 mol FeS 7 mol O = 0.30 mol FeS required Consider FeS : 0.9 g FeS x 1 mol FeS 87.9 g FeS = 0.38 mol FeS available 0.38 mol FeS x 7 mol O 4 mol FeS = mol O required therefore the limiting reagent is O mol O x mol Fe O 3 7 mol O x g Fe O 3 1 mol Fe O 3 = 18.4 g Fe O mol O x 4 mol SO 7 mol O x.414 L SO 1 mol SO = 5.16 L SO
19 7. Nickel metal can be highly purified using the Mond Process: Ni(s)+4CO(g) Ni(CO) 4 (g) In the first step of this process nickel metal is reacted with carbon monoxide under high pressure and heat to produce a gas product known as nickel carbonyl (Ni(CO) 4 ). If 40.0 g of nickel metal is reacted with 5.00 L of carbon monoxide at atm. pressure and a temperature of 875 K, calculate the resulting total pressure of all gases at 5 o C and total volume 5.00 L. Hints: nickel is the limiting reagent, Dalton s Law of Partial Pressures could be used to solve this problem Calculate the amount of CO(g) available: P = atm kpa x V = 5.00 L n =? R = kpa L K mol T = 875 K kpa 1 atm = 6155 kpa n = PV RT n = 6155 kpa 5.00 L kpa L 875 K K mol n = 4.31 mol CO available Calculate amount of CO(g) consumed in the reaction: 40.0 g Ni x 1 mol Ni 4 mol CO x g Ni 1 mol Ni Calculate the amount of CO remaining (unreacted): =.76 mol CO consumed amount CO remaining = (amount CO available) (amount CO consumed) = (4.31 mol CO) (.76 mol CO) = mol CO remains after reaction Calculate the amount of Ni(CO) 4 formed: 40.0 g Ni x 1 mol Ni g Ni x 1 mol Ni(CO) 4 1 mol Ni = mol Ni(CO) 4 formed Calculate the total amount of gases after reaction: total amount of gases = (amount CO remaining) (amount Ni(CO) 4 formed) = (1.505 mol CO) + ( mol Ni(CO) 4 ) =.186 mol of gas remains after reaction
20 Calculate the pressure of remaining gas: P =? V = 5.00 L n =.186 mol gas R = kpa L K mol T = 5 o C K P = nrt V P =.186 mol kpa L K mol 5.00 L P = 1084 kpa K 1084 kpa 1 atm kpa = atm
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