Chemistry 11 Solutios Sectio 9.2 Solutio Stoichiometry Solutios for Practice Problems Studet Editio page 417 11. Practice Problem (page 417) If 8.5 g of pure ammoium phosphate, (NH 4 ) 3 PO 4 (s), is dissolved i distilled water to make 400 ml of solutio, what are the cocetratios (i moles per litre) of the ios i the solutio? You eed to fid the molar cocetratio, c, of the ios i a solutio of ammoium phosphate. What Is Give? You kow the volume of the ammoium phosphate solutio: 400 ml You kow the mass of ammoium phosphate, (NH 4 ) 3 PO 4 (s): 8.5 g Pla Your Strategy Write the balaced chemical equatio for the dissolutio of ammoium phosphate, (NH 4 ) 3 PO 4 (s). Determie the molar mass of (NH 4 ) 3 PO 4 (s). Calculate the amout i moles of (NH 4 ) 3 PO 4 (s) usig the relatioship. M Covert the volume from millilitres to litres: 1 ml = 1 10 3 L Calculate the cocetratio of (NH 4 ) 3 PO 4 (aq) usig the relatioship c. Equate the mole ratios ad cross multiply to solve for, the amout i moles of (NH 4 ) 3 PO 4 (s). Act o Your Strategy Balaced equatio: (NH 4 ) 3 PO 4 (s) 3NH 4 + (aq) + PO 4 3 (aq) Mole ratio: 1 mole 3 moles 1 mole Molar mass, M, of (NH 4 ) 3 PO 4 (s): M 3 M 12 M 1 M 4M NH4 PO4 3 N H P O 3 14.01g/mol 12 1.01 g/mol 1 30.97 g / mol 4 16.00 g/mol 149.12 g/mol 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 18
Chemistry 11 Solutios Amout i moles,, of (NH 4 ) 3 PO 4 (s): m NH4 PO 3 4 M 8.5 g 149.12 g /mol 5.700 1 2 0 o m l Amout i moles,, of NH + 4 (aq): 1 mol (NH ) PO 0.005700 mol (NH ) PO 3 mol NH 4 3 4 4 3 4 4 NH4 NH4 3 mol NH 4 0.005700 mol (NH 4) PO 1 mol (NH 4) 3 PO4 0.0171 mol Amout i moles,, of PO 3 4 (aq): 1 mol (NH ) PO 0.005700 mol (NH ) PO 3 1 mol PO 4 3 4 4 3 4 4 3 PO4 3 PO4 3 4 1 mol PO 0.005700 mol (NH ) PO 0.005700 mol olume of solutio: 3 400 ml 1 10 L/ ml 0.400 L Cocetratio of NH + 4 ( aq): c 2 1.71 10 mol 0.400 L 0.04275 mol/l 0.04 mol/l 3 4 4 3 4 1 mol (NH 4) 3 PO4 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 19
Chemistry 11 Solutios Cocetratio of PO 3 4 (aq): c 2 0.005700 10 mol 0.400 L 0.01425 mol/l 0.01 mol/ L The cocetratio of the ammoium io, NH 4 + (aq), is 0.04 mol/l. The cocetratio of the phosphate io, PO 4 3 (aq), is 0.01 mol/l. Check Your Solutio The uits for amout ad cocetratio are correct. The aswer has oe sigificat digit ad seems reasoable. 12. Practice Problem (page 417) A strip of zic metal was placed i a beaker that cotaied 120 ml of a solutio of copper(ii) itrate, Cu(NO 3 ) 2 (aq). The mass of the copper produced was 0.813 g. Fid the iitial cocetratio of the copper(ii) itrate solutio. You eed to fid the molar cocetratio, c, of the copper(ii) itrate solutio. What Is Give? You kow the volume of the copper(ii) itrate solutio: 120 ml You kow the mass of copper precipitated: 0.813 g You kow the other reactat: zic Pla Your Strategy Write the chemical equatio for the sigle displacemet reactio. Use the periodic table to determie the atomic molar mass of Cu(s). Calculate the amout i moles of Cu(s) usig the relatioship. M Calculate the amout i moles of copper(ii) itrate usig the mole ratio i the balaced equatio. Covert the volume from millilitres to litres: 1 ml = 1 10 3 L Calculate the cocetratio of copper(ii) itrate usig the relatioship c. 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 20
Chemistry 11 Solutios Amout i moles,, of the precipitate, FeS(s): 1 mol FeS FeS 1 mol Fe(NO ) 0.0125 mol Fe(NO ) 3 2 3 2 FeS 1 mol FeS 0.0125 mol Fe(NO ) 1 mol Fe(NO 3 ) 2 0.0125 mol Molar mass, M, of the precipitate, FeS(s): M 1 M 1M FeS Fe S 1(55.85 g/mol) 1(32.07 g/mol) 87.92 g/mol 3 2 Mass, m, of FeS(s): m M FeS 0.0125 mol 1.099 g 1.10 g 87.92 g/ mol The precipitate is iro(ii) sulfide, FeS(s), ad the maximum mass that ca be collected from the reactio is 1.10 g. Check Your Solutio The uits for amout ad cocetratio are correct. The aswer has three sigificat digits ad seems reasoable. 15. Practice Problem (page 417) What mass of silver chloride, AgCl(s), ca be precipitated from 75 ml of 0.25 mol/l silver itrate, AgNO 3 (aq), by addig excess magesium chloride, MgCl 2 (aq)? You eed to calculate the mass of silver chloride that will precipitate i a reactio. What Is Give? You kow the volume of the silver itrate solutio: 75 ml You kow the iitial cocetratio of silver itrate: 0.25 mol/l You kow the other reactat is aqueous magesium chloride, MgCl 2 (aq). 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 25
Chemistry 11 Solutios Pla Your Strategy Predict the other product that forms i this double displacemet reactio. Write the balaced equatio for the reactio. Covert the volume to litres: 1 ml = 1 10 3 L Calculate the amout i moles of silver itrate usig the relatioship c. Equate the mole ratios ad solve for the amout i moles of precipitate. Use the periodic table to determie the molar mass of the precipitate. Calculate the mass of precipitate usig the relatioship m M. Act o Your Strategy The other product is magesium itrate, Mg(NO 3 ) 2 (aq). Balaced equatio: MgCl 2 (aq) + 2AgNO 3 (aq) Mg(NO 3 ) 2 (aq) + 2AgCl(s) Mole ratio: 1 mole 2 moles 1 mole 2 moles olume of solutio: 3 75 ml 1 10 L/ ml 0.075 L Amout i moles,, of AgNO 3 (aq): c AgNO 3 0.25 mol / L 0.075 L 0.01875 mol Amout i moles,, of precipitate, AgCl(s): 2 mol AgCl AgCl 2 mol AgNO 0.01875 mol AgNO AgCl 3 3 2 mol AgCl 0.01875 mol AgNO 2 mol AgNO 3 0.01875 mol Molar mass, M, of the precipitate, AgCl(s): M 1 M 1M AgCl Ag Cl 1 107.87 g/mol 1(35.45 g/mol) 143.32 g/mol 3 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 26
Chemistry 11 Solutios Mass, m, of AgCl(s): m M AgCl 0.01875 mol 2.687 g 2.7 g 143.32 g/ mol 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 27 The mass of silver chloride that ca be precipitated is 2.7 g. Check Your Solutio The uits for amout ad cocetratio are correct. The aswer has two sigificat digits ad seems reasoable. 16. Practice Problem (page 417) What mass of bromie gas ca be collected by bubblig excess chlorie gas through 850 ml of a 0.350 mol/l solutio of sodium bromide, NaBr(aq)? You eed to calculate the mass of bromie gas that will collect i a reactio. What Is Give? You kow the volume of the sodium bromide solutio: 850 ml You kow the iitial cocetratio of the sodium bromide solutio: 0.350 mol/l You kow the other reactat: chlorie gas, Cl 2 (aq) Pla Your Strategy Predict the ame ad formula for the other product that forms i this sigle displacemet reactio. Write the balaced equatio for the reactio. Covert the volume from millilitres to litres: 1 ml = 1 10 3 L Calculate the amout i moles of sodium bromide solutio usig the relatioship c. Equate the mole ratios ad solve for the amout i moles of bromie gas. Use the periodic table to determie the molar mass of the bromie gas, Br 2 (g). Calculate the mass of bromie gas usig the relatioship m M. Act o Your Strategy The other product that forms i this sigle displacemet reactio is sodium chloride, NaCl(aq). Balaced equatio: Cl 2 (g) + 2NaBr(aq) 2NaCl (aq) + Br 2 (g) Mole ratio: 1 mole 2 moles 2 moles 1 mole
Chemistry 11 Solutios Mass, m, of NaCl(s): m 0.213508 mol NaCl 12.477 g 12.5 g 58.44 g/ mol The mass of sodium chloride i the 1.0 L volumetric flask is 12.5 g. Check Your Solutio The uits for amout ad cocetratio are correct. The aswer has three sigificat digits ad seems reasoable. 20. Practice Problem (page 417) Food maufacturers sometimes add calcium acetate, Ca(CH 3 COO) 2 (s), to sauces as a thickeig aget. Whe aalyzed, a 250 ml solutio of calcium acetate was foud to cotai 0.200 mol of acetate ios. a. Fid the molar cocetratio of the calcium acetate solutio. b. What mass of calcium acetate was dissolved to make the solutio? a. molar cocetratio You eed to fid the molar cocetratio of a calcium acetate solutio. What Is Give? You kow the chemical formula for calcium acetate: Ca(CH 3 COO) 2 (s) You kow the amout i moles of acetate ios, CH 3 COO (aq): 0.200 mol You kow the volume of the solutio: 250 ml Pla Your Strategy Use the mole ratio of acetate ios to calcium acetate to determie the amout i moles of calcium acetate. Covert the volume from millilitres to litres: 1 ml = 1 10 3 L Calculate the cocetratio of calcium acetate usig the relatioship c. Act o Your Strategy The chemical formula for calcium acetate, Ca(CH 3 COO) 2 (s), idicates that there are two CH 3 COO ios for oe formula uit of Ca(CH 3 COO) 2 (s). 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 35
Chemistry 11 Solutios Amout i moles,, of Ca(CH 3 COO) 2 (s): 1 mol Ca(CH3COO) 2 Ca(CH3COO) 2 2 mol CH COO 0.200 mol CH COO 3 3 Ca(CH3COO) 2 olume of solutio: 250 ml 0.250 L 1 mol Ca(CH3COO) 2 0.200 mol CH3COO 2 mol CH COO 0.100 mol 3 1 10 L/ ml Cocetratio of Ca(CH 3 COO) 2 (aq): c 0.100 mol 0.250 L 0.40 mol/l The cocetratio of calcium acetate is 0.40 mol/l. b. mass of calcium acetate You eed to fid the mass of calcium acetate i 250 ml of solutio. What Is Give? You kow the molar cocetratio: 0.40 mol/l You kow the volume of solutio: 0.250 L Pla Your Strategy Calculate the amout i moles of calcium acetate usig the relatioship the relatioship c. Use the periodic table to fid the molar mass of Ca(CH 3 COO) 2 (s). Calculate the mass of calcium acetate usig the relatioship m M. 3 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 36
Chemistry 11 Solutios Act o Your Strategy Amout i moles,, of Ca(CH 3 COO) 2 (aq): CaCH3COO 2 c 0.40 mol/ L 0.250 L 0.10 mol Molar mass, M, of Ca(CH 3 COO) 2 (s): M 1 M 4 M 6 M 4M Ca CH COO 3 2 Ca C H O 1 40.08 g/mol 4 12.01 g/mol 6 1.01 g/mol 4 16.00 g/mol 158.18 g/mol Mass, m, of Ca(CH 3 COO) 2 (s): m M 3 2 Ca CH COO 0.10 mol 158.18 g/ mol 15.818 g 16 g The mass of calcium acetate is 16 g. Check Your Solutio The uits for amout ad cocetratio are correct. The aswer has two sigificat digits ad seems reasoable. Sectio 9.2 Solutio Stoichiometry Solutios for Practice Problems Studet Editio page 420 21. Practice Problem (page 420) Lead(II) sulfide, PbS(s), is a black, isoluble substace. Calculate the maximum mass of lead(ii) sulfide that will precipitate whe 6.75 g of sodium sulfide, Na 2 S(s), is added to 250 ml of 0.200 mol/l lead(ii) itrate, Pb(NO 3 ) 2 (aq). You eed to fid the mass of lead(ii) sulfide that will precipitate. 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 37