Section 7.4: Calculations Involving Limiting Reagents

Transcription

1 Sectio 7.4: Calculatios Ivolvig Limitig Reagets Tutorial 1 Practice, page Give:.3 mol ;.0 mol HNO = 3 Required: amout of water, NaCO = 3 O H Solutio: Step 1. List the give values ad the required value. HNO 3 (aq) + NaHCO 3 (s) H O(l) + CO (g) + NaNO 3 (aq).3 mol.0 mol H O Step. Determie the limitig reaget. NaHCO3 =.3 mol HNO3! 1 mol NaHCO 3 1 mol HNO3 NaHCO3 =.3 mol Sice the amout of sodium hydroge carboate preset iitially is less tha the required amout, sodium hydroge carboate is the limitig reaget. Step 3. Covert amout of sodium hydroge carboate to amout of water. H O =.0 mol NaHCO 3! 1 mol H O 1 mol NaHCO3 H O =.0 mol Statemet: Whe.3 mol of itric acid is combied with.0 mol of sodium hydroge carboate,.0 mol of water is produced.. Give: HCl = 5. mol ; MO = 1.5 mol Required: amout of chlorie, Cl Solutio: Step 1. List the give values ad the required value. 4 HCl(aq) + MO (s) Cl (g) + H O(l) + MCl (aq) 5. mol 1.5 mol Cl Step. Determie the limitig reaget. M = 5. mol HCl! 1 mol MO 4 mol HCl M = 1.3 mol Sice the amout of magaese dioxide preset is greater tha the required amout, hydrochloric acid is the limitig reaget. Copyright 011 Nelso Educatio Ltd. Chapter 7: Stoichiometry i Chemical Reactios 7.4-1

2 Step 3. Covert amout of hydrochloric acid to amout of chlorie. Cl = 5. mol HCl! 1 mol Cl 4 mol HCl Cl = 1.3 mol Statemet: 1.3 mol of chlorie ca be made from 5. mol of hydrochloric acid ad 1.5 mol of magaese dioxide. 3. (a) Al(s) + 3 I (s) AlI 3 (s) (b) Give: Al = 0.50 mol; I = 0.6 mol Required: amout of product, AlI3 Solutio: Step 1. List the give values ad the required value. Al(s) + 3 I (s) AlI 3 (s) 0.50 mol 0.60 mol AlI3 Step. Determie the limitig reaget. I = 0.50 mol Al! 3 mol I mol Al I = 0.75 mol Sice the amout of iodie preset is less tha the required amout, iodie is the limitig reaget. Step 3. Covert amout of iodie to amout of product. AlI3 = 0.60 mol I! mol AlI 3 3 mol I AlI3 = 0.4 mol Statemet: The amout of product that ca be made from 0.50 mol of alumium ad 0.60 mol of iodie is 0.4 mol. 4. (a) Al(s) + Fe O 3 (s) Fe(s) + Al O 3 (s) (b) Give: Al = 0.6 mol; Fe O 3 = 0.10 mol Required: amout of iro, Fe Solutio: Step 1. List the give values ad the required value. Al(s) + Fe O 3 (s) Fe(s) + Al O 3 (s) 0.6 mol 0.10 mol Fe Copyright 011 Nelso Educatio Ltd. Chapter 7: Stoichiometry i Chemical Reactios 7.4-

3 Step. Determie the limitig reaget. Fe O 3 = 0.6 mol Al! 1 mol Fe O 3 mol Al Fe O 3 = 0.13 mol Sice the amout of iro(iii) oxide preset is less tha the required amout, iro(iii) oxide is the limitig reaget. Step 3. Covert amout of iro(iii) oxide to amout of iro. Fe = 0.10 mol Fe O 3! mol Fe 1 mol Fe O 3 Fe = 0.0 mol Statemet: Whe 0.6 mol of alumium is combied with 0.10 mol of iro(iii) oxide, 0.0 mol of iro is expected. (c) Give: Al = 0.6 mol; Fe 0.10 mol O = 3 Required: amout of alumium oxide, Al O 3 Solutio: Step 1. List the give values ad the required value. Al(s) + Fe O 3 (s) Fe(s) + Al O 3 (s) 0.6 mol 0.10 mol Al O 3 Step. Covert amout of iro(iii) oxide to amout of alumium oxide. Al O 3 = 0.10 mol Fe O 3! 1 mol Al O 3 1 mol Fe O 3 Al O 3 = 0.0 mol Statemet: Whe 0.6 mol of alumium is combied with 0.10 mol of iro(iii) oxide, 0.10 mol of alumium oxide is expected. Tutorial Practice, page Give: m SiO = 10.0 g ; m g C = 7.00 Required: mass of silico carbide, m SiC Solutio: Step 1. List the give values, the required value, ad the correspodig molar masses. SiO (s) + 3 C(s) SiC(s) + CO (g) 10.0 g 7.00 g m SiC g/mol 1.01 g/mol g/mol Copyright 011 Nelso Educatio Ltd. Chapter 7: Stoichiometry i Chemical Reactios 7.4-3

4 Step. Covert mass of give substaces to amout of give substaces. Si = 10.0 g! 1 mol SiO g Si = mol [two extra digits carried] C = 7.00 g! 1 mol C 1.01 g C = mol [two extra digits carried] Step 3. Determie the limitig reaget. C = mol Si! 3 mol C 1 mol Si C = mol Sice the amout of carbo preset is greater tha the amout required, silico dioxide is the limitig reaget. Step 4. Covert amout of silico dioxide to amout of silico carbide. SiC = mol Si! 1 mol SiC 1 mol Si SiC = mol Step 5. Covert amout of silico carbide to mass of silico carbide.! g$ m SiC = ( mol )# & m SiC = 6.67 g Statemet: Whe 10.0 g of silico dioxide is combied with 7.00 g carbo, 6.67 g silico carbide is expected.. Give: m Fe = 5.00 g ; m Cl = 9.00 g Required: mass of iro(iii) chloride, m FeCl3 Solutio: Step 1. List the give values, the required value, ad the correspodig molar masses. Fe(s) + 3 Cl (g) FeCl 3 (s) 5.00 g 9.00g m FeCl g/mol g/mol 16.0 g/mol Step. Covert mass of give substaces to amout of give substaces. Fe = 5.00 g! 1 mol Fe g Fe = mol [two extra digits carried] Copyright 011 Nelso Educatio Ltd. Chapter 7: Stoichiometry i Chemical Reactios 7.4-4

5 Cl = 9.00 g! 1 mol Cl g Cl = mol [two extra digits carried] Step 3. Determie the limitig reaget. Cl = mol Fe! 3 mol Cl mol Fe Cl = mol Sice the amout of chlorie preset is less tha the required amout, chlorie is the limitig reaget. Step 4. Covert amout of chlorie to amout of iro(iii) chloride. FeCl3 = mol Cl! mol FeCl 3 3 mol Cl FeCl3 = mol Step 5. Covert amout of iro(iii) chloride to mass of iro(iii) chloride.! m = ( mol ) 16. g $ FeCl 3 # & mfecl 3 = 13.7 g Statemet: Whe 5.00 g of iro is combied with 9.00 g carbo, 13.7 g iro(iii) chloride is expected. 3. (a) Give: m 0.34g ; m 1.00 g NH = 3 O = Required: limitig reaget Solutio: Step 1. List the give values ad the correspodig molar masses. 4 NH 3 (g) + 5 O (g) 4 NO(g) + 6 H O(g) 0.34 g 1.00 g g/mol 3.00 g/mol Step. Covert mass of give substaces to amout of give substaces. NH3 = 0.34 g! 1 mol NH g NH3 = mol [two extra digits carried] = 1.00 g! 1 mol O 3.00 g = mol [two extra digits carried] Copyright 011 Nelso Educatio Ltd. Chapter 7: Stoichiometry i Chemical Reactios 7.4-5

6 Step 3. Determie the limitig reaget. = mol NH3! 5 mol O 4 mol NH3 = mol Statemet: Sice the amout of oxyge preset is greater tha the required amout, ammoia is the limitig reaget. (b) Give: m 0.34g ; m 1.00 g ; = mol NH 3 NH = 3 O = Required: m NO ; m H O Solutio: Step 1. List the give values, the required values, ad the correspodig molar masses. 4 NH 3 (g) + 5 O (g) 4 NO(g) + 6 H O(g) mol NO O 0.34 g 1.00 g m NO m H O g/mol 3.00 g/mol g/mol 18.0 g/mol Step. Covert amout of ammoia to amout of itroge mooxide. H NO = mol NH3! 4 mol NO 4 mol NH3 NO = mol Step 3. Covert amout of ammoia to amout of water. H O = mol NH 3! 6 mol H O 4 mol NH3 H O = mol Step 4. Covert amout of itroge mooxide to mass of itroge mooxide.! g$ m NO = ( mol )# & m NO = 0.60 g Step 5. Covert amout of water to mass of water.! 18.0 g$ m H = ( mol ) O # & m H O = 0.54 g Statemet: Whe 0.34 g of ammoia combies with 1.00 g of oxyge, 0.60 g of itroge mooxide ad 0.54 g water are produced i the reactio. Copyright 011 Nelso Educatio Ltd. Chapter 7: Stoichiometry i Chemical Reactios 7.4-6

7 Sectio 7.4 Questios, page Table 1 Amouts Ivolved i the Sythesis of Water H (g) + O (g) H O(g) Amout of hydroge (mol) Amout of oxyge (mol) Amout of water (mol) 1 mol O 6 4 mol H mol O mol H. Table Amouts Ivolved i the Sythesis of Ammoia N (g) + 3 H (g) NH 3 (g) Amout of itroge (mol) Amout of excess reaget remaiig (mol) Amout of hydroge (mol) Amout of ammoia (mol) Amout of excess reaget remaiig (mol) mol H mol N mol N mol H mol H 3. (a) Give: Mg = 0.58 mol ; 0.0 mol N = Required: limitig reaget; excess reaget Solutio: Step 1. List the give values. 3 Mg(s) + N (g) Mg 3 N (s) 0.58 mol 0.0 mol Step. Determie the limitig reaget. N = 0.58 mol Mg! 1 mol N 3 mol Mg N = 0.19 mol Statemet: Sice the amout of itroge preset is greater tha the required amout, magesium is the limitig reaget ad itroge is the excess reaget. (b) Give: Ca = 5.3 mol ; AlCl = 3.8 mol 3 Required: limitig reaget; excess reaget Solutio: Step 1. List the give values. 3 Ca(s) + AlCl 3 (aq) 3 CaCl (aq) + Al(s) 5.3 mol 3.8 mol Step. Determie the limitig reaget. AlCl3 = 5.3 mol Ca! mol AlCl 3 3 mol Ca AlCl3 = 3.5 mol Statemet: Sice the amout of alumium chloride preset is greater tha the amout required to react with calcium, calcium is the limitig reaget ad alumium chloride is the excess reaget. Copyright 011 Nelso Educatio Ltd. Chapter 7: Stoichiometry i Chemical Reactios 7.4-7

8 (c) Give: 0.10 mol ; 0.35 mol FeS = O = Required: limitig reaget; excess reaget Solutio: Step 1. List the give values. 4 FeS (s) + 11 O (g) Fe O 3 (s) + 8 SO (g) 0.10 mol 0.35 mol Step. Determie the limitig reaget. = 0.10 mol FeS! 11 mol O 4 mol FeS = 0.8 mol Statemet: Sice the amout of oxyge preset is greater tha the required amout, iro pyrite is the limitig reaget ad oxyge is the excess reaget. 4. (a) Give: Cu = 0.4 mol ; AgNO = 0.5 mol 3 Required: amout of silver, Ag Solutio: Step 1. List the give values ad the required value. Cu(s) + AgNO 3 (aq) Ag(s) + Cu(NO 3 ) (aq) 0.4 mol 0.5 mol Ag Step. Determie the limitig reaget. AgNO3 = 0.4 mol Cu! mol AgNO 3 1 mol Cu AgNO3 = 0.48 mol Sice the amout of silver itrate preset is greater tha the amout required to react with copper, copper is the limitig reaget. Step 3. Covert amout of copper to amout of silver. Ag = 0.4 mol Cu! mol Ag 1 mol Cu Ag = 0.48 mol Statemet: If 0.4 mol of copper were combied with 0.5 mol of silver itrate, 0.48 mol of silver would be produced. (b) Silver itrate is i excess. amout of silver itrate preset iitially = 0.5 mol amout of silver itrate required = 0.48 mol 0.5 mol 0.48 mol = 0.04 mol Therefore, the amout of silver itrate remaiig would be 0.04 mol. Copyright 011 Nelso Educatio Ltd. Chapter 7: Stoichiometry i Chemical Reactios 7.4-8

9 5. (a) Give: Al = 0.35 mol ; HCl = 1. mol Required: amout of alumium chloride, AlCl3 Solutio: Step 1. List the give values ad the required value. Al(s) + 6 HCl(aq) AlCl 3 (aq) + 3 H (g) 0.35 mol 1. mol Step. Determie the limitig reaget. HCl = 0.35 mol Al! 6 mol HCl mol Al HCl = 1.05 mol Sice the amout of hydrochloric acid preset is greater tha the amout required to react with alumium, alumium is the limitig reaget. Step 3. Covert amout of alumium to amout of alumium chloride. AlCl3 = 0.35 mol Al! mol AlCl 3 mol Al AlCl3 = 0.35 mol Statemet: Whe 0.35 mol of alumium is combied with 1. mol of hydrochloric acid, 0.35 mol of alumium chloride will be produced. (b) Hydrochloric acid is i excess. amout of hydrochloric acid preset iitially = 1. mol amout of hydrochloric acid required = 1.05 mol 1. mol 1.05 mol = 0.15 mol Therefore, the amout of hydrochloric acid remaiig will be 0.15 mol. 6. Give: 5.8 mol ;.8 mol SO = O = Required: mass of sulfur trioxide, m SO3 Solutio: Step 1. List the give values, the required value, ad the correspodig molar masses. SO (g) + O (g) SO 3 (g) 5.8 mol.8 mol m SO g/mol 3.00 g/mol g/mol Step. Determie the limitig reaget. = 5.8 mol S! 1 mol O mol S =.9 mol Sice the amout of oxyge preset is less tha the amout required to react with sulfur dioxide, oxyge is the limitig reaget. Copyright 011 Nelso Educatio Ltd. Chapter 7: Stoichiometry i Chemical Reactios 7.4-9

10 Step 3. Covert amout of oxyge to amout of sulfur trioxide. SO3 =.8 mol! mol SO 3 1 mol SO3 = 5.6 mol Step 4. Covert amout of sulfur trioxide to mass of sulfur trioxide.! g$ m SO3 = (5.6 mol) # & m SO3 = 450 g Statemet: Whe 5.8 mol of sulfur dioxide ad.8 mol of oxyge are combied, 450 g of sulfur trioxide will be produced. 7. (a) H (g) + Cl (g) HCl(g) (b) Give: m 10.0 g ; m 30.0 g H = Cl = Required: mass of hydroge chloride, m HCl Solutio: Step 1. List the give values, the required value, ad the correspodig molar masses. H (g) + Cl (g) HCl(g) 10.0 g 30.0 g m HCl.0 g/mol g/mol g/mol Step. Covert mass of give substaces to amout of give substaces. H = 10.0 g! 1 mol H.0 g H = mol [two extra digits carried] Cl = 30.0 g! 1 mol Cl g Cl = mol [two extra digits carried] Step 3. Determie the limitig reaget. The mole ratio of hydroge to chlorie is 1:1, the amout of chlorie preset is less tha the amout of chlorie required to react with hydroge. Therefore, chlorie is the limitig reaget. Step 4. Covert amout of chlorie to amout of hydroge chloride. HCl = mol Cl! mol HCl 1 mol Cl HCl = mol [two extra digits carried] Copyright 011 Nelso Educatio Ltd. Chapter 7: Stoichiometry i Chemical Reactios

11 Step 5. Covert amout of hydroge chloride to mass of hydroge chloride.! g$ m HCl = (9.068 mol )# & m HCl = 39 g Statemet: If 10.0 g of hydroge mixes with 30.0 g of chlorie, 39 g of hydroge chloride will be produced. 8. Give: m Al(OH) = 0.50 g ; m g 3 HCl = 0.60 Required: mass of alumium chloride, m AlCl3 Solutio: Step 1. List the give values, the required value, ad the correspodig molar masses. Al(OH) 3 (s) + 3 HCl(aq) 3 H O(l) + AlCl 3 (aq) 0.50 g 0.60 g m AlCl g/mol g/mol g/mol Step. Covert mass of give substaces to amout of give substaces. Al(OH)3 = 0.50 g! 1 mol Al(OH) g Al(OH)3 = mol [two extra digits carried] HCl = 0.60 g! 1 mol HCl g HCl = mol [two extra digits carried] Step 3. Determie the limitig reaget. HCl = mol Al(OH)3! 3 mol HCl 1 mol Al(OH)3 HCl = mol [two extra digits carried] Sice the amout of hydrochloric acid preset is less tha the amout required to react with alumium hydroxide, hydrochloric acid is the limitig reaget. Step 4. Covert amout of hydrochloric acid to amout of alumium chloride. AlCl3 = mol HCl! 1 mol AlCl 3 3 mol HCl AlCl3 = mol [two extra digits carried] Copyright 011 Nelso Educatio Ltd. Chapter 7: Stoichiometry i Chemical Reactios

12 Step 5. Covert amout of alumium chloride to mass of alumium chloride.! m AlCl3 = ( mol ) g $ # & m AlCl3 = 0.73 g Statemet: If 0.50 g of alumium hydroxide is placed i a solutio cotaiig 0.60 g of hydrochloric acid, 0.73 g of alumium chloride will be produced. 9. Give: m 10.0 g ; m 30.0 g C 4H 10 = O = Required: mass of carbo dioxide, m C Solutio: Step 1. List the give values, the required value, ad the correspodig molar masses. C 4 H 10 (g) + 13 O (g) 8 CO (g) + 10 H O(g) 10.0 g 30.0 g m C g/mol 3.00 g/mol g/mol Step. Covert mass of give substaces to amout of give substaces. C4 H 10 = 10.0 g! 1 mol C 4 H g C4 H 10 = mol [two extra digits carried] = 30.0 g! 1 mol O 3.00!!!! g = mol [two extra digits carried] Step 3. Determie the limitig reaget. = mol C4 H 10! 13 mol O mol C4 H 10 = mol [two extra digits carried] Sice the amout of oxyge preset is less tha the amout required to react with butae, oxyge is the limitig reaget. Step 4. Covert amout of oxyge to amout of carbo dioxide. C = mol! 8 mol CO 13 mol C = mol [two extra digits carried] Copyright 011 Nelso Educatio Ltd. Chapter 7: Stoichiometry i Chemical Reactios 7.4-1

13 Step 5. Covert amout of carbo dioxide to mass of carbo dioxide.! g$ m C = ( mol )# & m C = 5.4 g Statemet: The mass of carbo dioxide produced from the combustio of 10.0 g of butae ad 30.0 g of oxyge is 5.4 g. 10. (a) Give: m TiO = 40.0 g ; m g C = 7.0 ; m Cl = 30.0 g Required: limitig reaget Solutio: Step 1. List the give values ad the correspodig molar masses. TiO (s) + C(s) + Cl (g) TiCl 4 (g) + CO (g) 40.0 g 7.0 g 30.0 g g/mol 1.01 g/mol g/mol Step. Covert mass of give substaces to amout of give substaces. Ti = 40.0 g! 1 mol TiO g Ti = mol [two extra digits carried] Cl = 30.0 g! 1 mol Cl g Cl = mol [two extra digits carried] C = 7.0 g! 1 mol C 1.01 g C = mol [two extra digits carried] Statemet: Accordig to the balaced chemical equatio, the mole ratio of titaium(iv) oxide to carbo to chlorie is 1:1:. The amout of chlorie required is twice the amout of titaium or the amout of carbo. Sice the amout of chlorie is less tha the miimum of these two amouts, chlorie is the limitig reaget. (b) Give: mol Cl = Required: mass of titaium(iv) chloride, m TiCl4 Solutio: Step 1. List the give value, the required value, ad the correspodig molar masses. TiO (s) + C(s) + Cl (g) TiCl 4 (g) + CO (g) mol m TiCl g/mol g/mol Copyright 011 Nelso Educatio Ltd. Chapter 7: Stoichiometry i Chemical Reactios

14 Step. Covert amout of chlorie to amout of titaium(iv) chloride. TiCl4 = mol Cl! 1 mol TiCl 4 mol Cl TiCl4 = mol [two extra digits carried] Step 3. Covert amout of titaium(iv) chloride to mass of titaium(iv) chloride.! m TiO4 = ( mol ) g $ # & m TiO4 = 40.1 g Statemet: The mass of titaium(iv) chloride that ca be produced from mixig 40.0 g of titaium(iv) oxide, 7.0 g of carbo, ad 30.0 g of chlorie is 40.1 g. 11. Give: m 15.0 g ; m 6.0 g NH = 3 CH = 4 Required: mass of hydroge cyaide, m HCN Aalysis: Use the equatios to determie the mole ratio of ammoia to methae i the process. 4 NH 3 (g) + 5 O (g) 4 NO(g) + 6 H O(g) amout of NH 3 to amout of NO = 4:4 = 1:1 NO(g) + CH 4 (g) HCN(g) + H O(g) + H (g) amout of NO to amout of CH 4 = : = 1:1 Therefore, the amouts of NH 3 ad CH 4 are i the ratio 1:1. The, determie the mass of hydroge cyaide produced from the give masses of ammoia ad methae. Solutio: Step 1. List the give values, the required value, ad the correspodig molar masses. 4 NH 3 (g) + 5 O (g) 4 NO(g) + 6 H O(g) 15.0 g g/mol NO(g) + CH 4 (g) HCN(g) + H O(g) + H (g) 6.0 g m HCN g/mol 7.03 g/mol Step. Covert mass of give substaces to amout of give substaces. NH3 = 15.0 g! 1 mol NH g NH3 = mol [two extra digits carried] CH4 = 6.0 g! 1 mol CH g CH4 = mol [two extra digits carried] Copyright 011 Nelso Educatio Ltd. Chapter 7: Stoichiometry i Chemical Reactios

15 Step 3. Determie the limitig reaget. CH4 = mol NH3! 1 mol CH 4 1 mol NH3 CH4 = mol [two extra digits carried] Sice the amout of methae preset is less tha the required amout, methae is the limitig reaget. Step 4. Covert amout of methae to amout of hydroge cyaide. HCN = mol CH4! mol HCN mol CH4 HCN = mol [two extra digits carried] Step 5. Covert amout of hydroge cyaide to mass of hydroge cyaide.! 7.03 g$ m HCN = ( mol )# & m HCN = 10 g Statemet: If 15.0 g of ammoia ad 6.0 g of methae are preset iitially, 10 g of hydroge cyaide will be produced. Copyright 011 Nelso Educatio Ltd. Chapter 7: Stoichiometry i Chemical Reactios