Act o Your Strategy Amout i moles,, of Ca(CH 3 COO) 2 (aq): CaCH3COO 2 c V 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). What Is Required? You eed to fid the mass of lead(ii) sulfide that will precipitate. 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 37
What Is Give? You kow the mass of the sodium sulfide solutio: 6.75 g You kow the volume of the lead(ii) itrate solutio: 250 ml You kow the cocetratio of the lead(ii) itrate solutio: 0.200 mol/l Pla Your Strategy Write the balaced chemical equatio for the reactio. Determie the molar mass of Na 2 S(s). Calculate the amout i moles of each reactat. To allow for the mole ratio of the reactats, divide the amout of each reactat by its coefficiet i the chemical equatio. The smaller result idetifies the limitig reactat. Use the mole ratio i the balaced equatio ad the amout i moles of the limitig reactat to fid the amout i moles of the precipitate. Determie the molar mass of PbS(s). Calculate the mass of PbS(s) usig the relatioship m M. Act o Your Strategy Balaced chemical equatio: Na 2 S(s) + Pb(NO 3 ) 2 (aq) 2NaNO 3 (aq) + PbS(s) Amout i moles,, of Na 2 S(s): Na2S M 6.75 g 78.05 g /mol 8.6483 1 0 o m l Molar mass, M, of Na 2 S(s): M 2M 1M Na2S Na S 2 22.99 g/mol +1 32.07 g/mol 78.05 g/mol Amout i moles,, of Pb(NO 3 ) 2 (aq): c V Pb NO 3 2 0.200 mol/ L 0.250 L 5.00 10 mol 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 38
Idetificatio of the limitig reactat: amout of Na 2S 8.6483 10 mol coefficiet 1 8.6483 10 mol 2 amout of Pb(NO 3) 2 5. 00 10 mol coefficiet 1 2 5.00 10 mo Pb(NO 3 ) 2 (aq) is the limitig reactat because it is the smaller amout. Amout i moles,, of the precipitate, PbS(s): 1 mol Pb(NO ) 5.00 10 mol Pb(NO ) 1 mol PbS 3 2 3 2 PbS Molar mass, M, of PbS(s): M 1 M 1M PbS Pb S PbS 1 mol PbS 5.00 10 mol Pb(NO 3) 2 5.00 10 mol 1 mol Pb(NO 3) 2 1 207.2 g/mol + 1 32.07 g/mol 239.27 g / mol Mass, m, of PbS(s): m M PbS 239.27 g/ mol 5.00 10 mol 11.963 g 12 g The mass of lead(ii) sulfide that precipitates is 12 g. Check Your Solutio The mass of precipitate seems reasoable compared with the umber of moles of reactat used i this reactio. The aswer correctly shows two sigificat digits. l 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 39
22. Practice Problem (page 420) Silver chromate, Ag 2 CrO 4 (s), is a brick-red isoluble substace that is used to stai euros so that they ca be viewed uder a microscope. Silver chromate ca be formed by the reactio betwee silver itrate, AgNO 3 (aq), ad potassium chromate, K 2 CrO 4 (aq), as show i the photograph below. Calculate the mass of silver chromate that forms whe 25.0 ml of 0.125 mol/l silver itrate reacts with 20.0 ml of 0.150 mol/l potassium chromate. What Is Required? You eed to fid the mass of silver chromate that will precipitate. What Is Give? You kow the volume of the silver itrate solutio: 25.0 ml You kow the cocetratio of the silver itrate solutio: 0.125 mol/l You kow the volume of the potassium chromate solutio: 20.0 ml You kow the cocetratio of the potassium chromate solutio: 0.150 mol/l Pla Your Strategy Write the balaced chemical equatio for the reactio. Calculate the amout i moles of each reactat usig the relatioship c V. To allow for the mole ratio of the reactats, divide the amout of each reactat by its coefficiet i the chemical equatio. The smaller result idetifies the limitig reactat. Use the mole ratios to fid the amout i moles,, of the precipitate. Determie the molar mass of Ag 2 CrO 4 (s). Calculate the mass of Ag 2 CrO 4 (s) usig the relatioship m M. 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 40
Amout i moles,, of Na 2 CO 3 (aq): 1 mol Ba 6.250 10 mol Ba 1 mol Na CO 2 2 2 3 Na2CO3 Na2CO3 1 mol Na 2CO 3 6.250 10 mol Ba 2 1 mol Ba 6.250 10 mol Molar mass, M, of Na 2 CO 3 (s): M 2 M 1 M 3M Na2CO3 Na C O 2 2 22.99 g/mol +1 12.01 g/mol +3 16.00 g/mol 105.99 g/mol Mass, m, of Na 2 CO 3 (s): m M Na2CO3 6.25 10 mol 1 6.6254 10 g 1 6.62 10 g 105.99 g/ mol The mass of sodium carboate required is 6.62 10 1 g. Check Your Solutio The mass is correctly expressed i grams ad shows three sigificat digits. This aswer seems reasoable based upo the mole ratio i the balaced equatio ad the quatity of reactat that has bee give. 25. Practice Problem (page 420) What is the maximum mass of lead(ii) iodide, PbI 2 (s), that ca precipitate whe 40.0 ml of a 0.345 mol/l solutio of lead(ii) itrate, Pb(NO 3 ) 2 (aq), is mixed with 85.0 ml of a 0.210 mol/l solutio of potassium iodide, KI(aq)? Why might the actual mass precipitated be less? What Is Required? You eed to the fid the maximum mass of lead iodide that will precipitate whe solutios of lead(ii) itrate ad potassium iodide are mixed. 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 45
What Is Give? You kow the volume of the potassium iodide solutio: 85.0 ml You kow the cocetratio of the potassium iodide solutio: 0.210 mol/l You kow the volume of the lead(ii) itrate solutio: 40.0 ml You kow the cocetratio of the lead(ii) itrate solutio: 0.345 mol/l Pla Your Strategy Write the balaced chemical equatio for the reactio. Calculate the amout i moles of each reactat usig the relatioship the relatioship c V. To allow for the mole ratio of the reactats, divide the amout of each reactat by its coefficiet i the chemical equatio. The smaller result idetifies the limitig reactat. Use the mole ratio i the balaced equatio ad the amout i moles of the limitig reactat to fid the amout i moles of the precipitate, PbI 2 (s). Determie the molar mass of PbI 2 (s). Calculate the mass of PbI 2 (s) usig the relatioship m M. Act o Your Strategy Balaced chemical equatio: 2KI(aq) + Pb(NO 3 ) 2 (aq) 2KNO 3 (aq) + PbI 2 (s) Amout i moles,, of KI(aq): c V KI 0.210 mol/ L 0.0850 L 1.785 10 mol Amout i moles,, of Pb(NO 3 ) 2 (aq): c V Pb NO 3 2 0.345 mol/ L 0.0400 L 1.380 10 mol Idetificatio of limitig reactat: amout of KI 1.785 10 mol coefficiet 2 8.925 10 mol 2 amout of Pb(NO 3) 2 1.380 10 mol coefficiet 1 2 1.380 10 mol 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 46
KI(aq) is the limitig reactat because it is the smaller amout. Amout i moles,, of the precipitate, PbI 2 (s): 2 mol KI 1.78510 mol KI 1 mol PbI PbI2 2 PbI2 1 mol PbI 2 1.785 10 mol KI 8.925 10 mol Molar mass, M, of PbI 2 (s): M 1 M 2M PbI2 Pb I 2 mol KI 1 207.2 g/mol +2 126.90 g/mol 461.00 g / mol Mass, m, of PbI 2 (s): m M PbI 2 8.925 10 mol 4.1144 g 4.11 g 461.00 g/ mol The maximum mass of lead(ii) iodide that precipitates is 4.11 g. Oe reaso that the mass of PbI 2 (s) could be less tha this amout is that a small amout of solid lead iodide dissolves. I additio, if the mass of precipitate was obtaied by filterig, some of the precipitate could pass through the filter paper with the filtrate, resultig i a lower recovered mass of precipitate. Check Your Solutio The mass of precipitate seems reasoable compared with the amout i moles of the reactat used i this reactio. The aswer correctly shows three sigificat digits. 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 47
Amout i moles,, of the precipitate, CaCO 3 (s): 1 mol CaCl2 5.000 10 mol CaCl 2 1 mol CaCO 3 CaCO3 CaCO3 1 mol CaCO 5.000 10 mol CaCl 1 mol CaCl 2 5.000 10 mol Molar mass, M, of CaCO 3 (s): M 1 M 1 M 3M CaCO3 Ca C O 3 2 1 40.08 g/mol +1 12.01 g/ mol +3 16.00 g/mol 100.09 g / mol Mass, m, of CaCO 3 (s): m M CaCO 3 5.000 10 mol 0.500 g 100.09 g/ mol The mass of calcium carboate that precipitates is 0.500 g. Check Your Solutio The mass of precipitate seems reasoable compared with the amout i moles of reactat used. The aswer correctly shows three sigificat digits. 30. Practice Problem (page 420) Barium chromate, BaCrO 4 (s), is a isoluble yellow solid. Determie the cocetratio of barium ios i a solutio made by mixig 50.0 ml of a 0.150 mol/l solutio of barium itrate, Ba(NO 3 ) 2 (aq), with 50.0 ml of a 0.120 mol/l solutio of potassium chromate, K 2 CrO 4 (aq). What Is Required? You eed to fid the cocetratio of barium ios remaiig i a solutio. What Is Give? You kow the volume of the barium itrate solutio: 50.0 ml You kow the cocetratio of the barium itrate solutio: 0.150 mol/l You kow the volume of the potassium chromate solutio: 50.0 ml You kow the cocetratio of the potassium chromate solutio: 0.120 mol/l 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 54
Pla Your Strategy Write the balaced chemical equatio for the reactio. Calculate the amout i moles of each reactat usig the relatioship c V. To allow for the mole ratio of the reactats, divide the amout of each reactat by its coefficiet i the chemical equatio. The smaller result idetifies the limitig reactat. Determie the amout i moles of Ba(NO 3 ) 2 (aq) i excess. Determie the amout i moles of Ba 2+ (aq) per mole of Ba(NO 3 ) 2 (aq). Determie the amout i moles of Ba 2+ i excess. Calculate the total volume of the mixture ad determie the cocetratio of Ba 2+ (aq). Act o Your Strategy Balaced equatio: Ba(NO 3 ) 2 (aq) + K 2 CrO 4 (aq) 2KNO 3 (aq) + BaCrO 4 (s) Mole ratio 1 mole 1 mole 2 mole 1 mole Amout i moles,, of Ba(NO 3 ) 2 (aq): c V Ba NO 3 2 0.150 mol/ L 0.0500 L 7.50 10 mol Amout i moles,, of K 2 CrO 4 (aq): c V K2CrO 4 0.120 mol/ L 0.0500 L 6.00 10 mol = 0.120 mol/ L 0.0500 L Idetificatio of the limitig reactat: amout of Ba(NO 3) 2 7.50 10 mol coefficiet 1 7. 50 10 mol amout of K2CrO 4 6.00 10 mol coefficiet 1 6.00 10 mo Ba(NO 3 ) 2 (aq) is the limitig reactat because it is the smaller amout. l 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 55
Sice the mole ratio of Ba(NO 3 ) 2 (aq) to K 2 CrO 4 (aq) is 1:1, the excess amout i moles,, of Ba(NO 3 ) 2 (aq) is the differece betwee the amout i moles of the two reactats. BaNO excess Ba NO3 2 3 2 K2CrO4 7.50 10 mol 6.00 10 mol 1.5 10 mol Amout i moles,, of Ba 2+ (aq): 2 2 Ba 1 mol Ba 1.50 10 mol Ba NO 1 mol Ba NO 3 2 3 2 2 Ba 1.50 10 mol Ba NO3 2 1.50 10 o 1 mol Ba(NO m l ) 3 2 1 mol Ba 2 Total volume of mixture: V = 50.0 ml + 50.0 ml = 100.0 ml = 0.100 L Cocetratio of Ba 2+ (aq): c V 1.50 10 mol 0.100 L 1.50 10 mol/l The cocetratio of barium ios remaiig i solutio is 1.50 10 mol/l. Check Your Solutio The uits i the calculatios have cacelled properly ad the fial uit is correct. The cocetratio of the barium io i excess seems reasoable. The aswer correctly shows three sigificat digits. 9780071051071Chapter 9ReactiosiAqueousSolutios MHR 56