P B.1 (pg 1 of 5) Solution Stoichiometry Name

Transcription

1 P B.1 (pg 1 of 5) Solution Stoichiometry Name Write out a balanced equation that represents the combination of the following aqueous salt, and then complete the stoichiometry problem. Write both the molecular equation and the net ionic equation. 1. If 25.0 ml of a 0.4 M solution of lead(ii) nitrate is combined with 25.0 ml of 0.4 M sodium iodide solution, a. calculate the mass of the precipitate that should form. b. If 1.50 g of precipitate did form in the laboratory, calculate the % yield. 2. If 45.0 ml of a 0.54 M solution of barium nitrate is combined with 5.0 ml of 0.54 M sodium phosphate solution, a. calculate the mass of the precipitate that should form. b. If.2 g of precipitate did form in the laboratory, calculate the % yield.. If 40.0 ml of 0.87 M solution of silver nitrate is combined with 55.0 ml of 0.57 M of potassium chromate, a. calculate the mass of precipitate that should form. b. If 4.96 g of the precipitate forms in the laboratory, calculate the % yield of ppt formed. 4. If an excess of cobalt (II) chloride is combined with 5.0 ml of a potassium carbonate solution, and 8.11 g of the precipitate forms, determine the original molarity of the potassium carbonate solution. 5. If 15.0 ml of potassium chromate is combined with an excess of a aluminum chloride solution, and 1.54 g of the precipitate forms, determine the original molarity of the potassium chromate solution. 6. If a volume of 0.50 M cobalt (II) chloride is combined with an excess of potassium ferrocyanide solution, and 4.7 g of the precipitate forms, determine the original volume of the cobalt (II) chloride solution. NOTE: ferrocyanide ion is Fe(CN) If a 0.64 M solution of copper (II) sulfate solution is combined with an excess of potassium chromate, and 12. g of the precipitate forms, determine the original volume of the copper (II) sulfate solution used. 8. If an excess of cobalt (II) chloride is combined with a 0.5 M solution of potassium chromate, and.86 g of the precipitate forms, determine the original volume of the potassium chromate solution used.

2 P B.1 (pg 2 of 5) Solution Stoichiometry ANSWERS 1. molecular: Pb(NO)2 + 2 NaI PbI2 (ppt) + 2 NaNO net ionic Pb I PbI2 (ppt) First, be sure you are working in moles change both reactants to mole quantities (0.025 L)(0.4 M) = moles Pb(NO)2 (0.025 L)(0.4 M) = moles of NaI a. determine that the sodium iodide limits then calculate the mass of ppt that can form molNaI 1PbI 2 2NaI = moles of PbI2 (ppt) can be formed 461g 1mol = 1.96 g of ppt theor b. then determine the % yield 1.50gExperimental 1.96gTheoretical 100 = 76.6% First, realize that there will be no iodide ions left in solution since they are the limiting reactant. 2NO molPb(NO 1Pb(NO = moles 0.017molesNO NO, next calculate molarity 0.05L(totalVol) = 0.4 M NO molNaI 1Na+ 1NaI = moles Na+ ions, next calculate molarity Na L(totalVol) = M Na+ Then determine the moles of lead that are needed to go with the limiting reactant by doing stoichiometry molNaI 1Pb(NO ) 2 2NaI = moles of Pb(NO)2 required to go with all the I ions then calculate the amount of lead ion that will be left over (Lead is the excess reactant so some of it will remain.) Thus ( moles of Pb 2+ started with) ( moles of Pb 2+ needed) = moles Pb 2+ left over and then calculate Pb 2+ molarity 0.05L(totalVol) = M Pb2+ ions remaining in solution. 2. Ba(NO)2 + 2 NaPO4 Ba(PO4)2 (ppt) + 6 NaNO NET: Ba PO4 Ba(PO4)2 (ppt) a g ppt b % yield c. no barium ions left over, 0.61 M nitrate ions, 0.71 M sodium ions, 0.04 M phosphate ions. 2 AgNO + K2CrO4 Ag2CrO4(ppt) + 2 KNO NET: 2 Ag + + CrO4 2 Ag2CrO4(ppt) a g ppt b % yield c. no silver ions left over, 0.7 M nitrate ions, 0.66 M potassium ions, 0.15 M chromate ions M original solution M original solution L of original solution L of original solution L of original solution Please go to P B.1 on unit B document page for worked out answers to the rest of these problems.

3 P B.1 (pg of 5) Solution Stoichiometry ANSWERS 2. molecular: Ba(NO)2 + 2 NaPO4 Ba(PO4)2 (ppt) + 6 NaNO net ionic: Ba PO4 Ba(PO4)2 (ppt) change both reactants to mole quantities (0.045 L)(0.54 M) = moles Ba(NO)2 (0.05 L)(0.54 M) = moles of NaPO4 a. determine that the barium nitrate limits then calculate the mass of ppt that can form 0.024molBa(NO 1Ba (PO ) 4 2 Ba(NO = moles of Ba(PO4)2 (ppt) can be formed 602g 1mol = 4.88 g of ppt b. determine the % yield.2gexperimental 4.88gTheoretical 100 = 68.1% first, realize that there will be no barium ions left in solution since they are the limiting reactant. 2NO 0.024molBa(NO 1Ba(NO = moles NO, next calculate molarity molesNO 0.08L(totalVol) = 0.61 M NO Na molNa PO 4 1Na PO 4 = moles Na+ ions, next calculate molarity Na L(totalVol) = 0.71 M Na+ Then determine the moles of phosphate that are needed to go with the limiting reactant by doing stoichiometry 2Na 0.024molBa(NO PO 4 Ba(NO = moles of PO4 required to go with all the Ba 2+ then calculate the amount of phosphate ion that will be left over (Phosphate is the excess reactant so some of it will remain.) ( moles of PO4 started with) ( moles of PO4 needed) = mol PO4 left over and then calculate the molPO molarity L(totalVol) = 0.04 M PO4

4 P B.1 (pg 4 of 5) Solution Stoichiometry ANSWERS. molecular: 2 AgNO + K2CrO4 Ag2CrO4(ppt) + 2 KNO net ionic: 2 Ag + + CrO4 2 Ag2CrO4(ppt) change both reactants to mole quantities (0.040 L)(0.87 M) = moles AgNO (0.055 L)(0.57 M) = moles of K2CrO4 a. determine that the silver nitrate limits then calculate the mass of ppt that can form 0.048molAgNO 1Ag CrO 1.8g 2AgNO = moles of Ag2CrO4(ppt) can be formed 1mol = 5.77 g of ppt b. determine the % yield 4.96gExperimental 5.77gTheoretical 100 = 86.0% first, realize that there will be no silver ions left in solution since they are the limiting reactant. 1NO 0.048molAgNO 1AgNO = moles 0.048molesNO NO, next calculate molarity 0.095L(totalVol) = 0.7 M NO 2K molK 2 1K 2 = moles K+ ions, next calculate molarity K L(totalVol) = 0.66 M K+ Then determine the moles of chromate that are needed to go with the limiting reactant by doing stoichiometry 0.048molAgNO 1K CrO 2AgNO = moles of CrO42 required to go with all the Ag + then calculate the amount of chromate ion that will be left over (Chromate is the excess reactant so some of it will remain.) (0.015 moles of CrO4 2 started with) ( moles of CrO4 2 needed) = moles CrO4 2 left over and then molCrO calculate the molarity L(totalVol) = 0.15M CrO42 4. molecular: CoCl2 + K2CO CoCO (ppt) + 2 KCl net ionic Co 2+ + CO 2 CoCO (ppt) the mass of precipitate will determine the original molarity of the potassium chromate solution 1mol 8.11g 118.9g = moles ppt which is the same moles of K2CO because of the 1:1 ratio molK 2 CO 0.05L(totalVol) = 1.95 M of K2CO solution

5 P B.1 (pg 5 of 5) Solution Stoichiometry ANSWERS 5. molecular: 2 AlCl + K2CrO4 Al2(CrO4) (ppt) + 6 KCl net ionic: 2 Al + + CrO4 2 Al2(CrO4) (ppt) the mass of precipitate will determine the original molarity of the potassium chromate solution 1.54g 1mol 402g = moles ppt K 2 1Al 2 ( ) = moles of K2CrO molK L(totalVol) = 0.77 M of K2CrO4 solution 6. molecular: 2 CoCl2 + K4Fe(CN)6 Co2Fe(CN)6 (ppt) + 4 KCl net ionic: 2 Co 2+ + Fe(CN)6 4- Co2Fe(CN)6 (ppt) the mass of precipitate will determine the original volume of the cobalt(ii) chloride solution 4.7g 1mol 1g = moles ppt 2CoCl 2 1Co 2 Fe(CN) 6 = moles of CoCl molCoCl M = L of CoCl2 solution, which is 52.8 ml CoCl2 solution 7. molecular: CuSO4 + K2CrO4 CuCrO4 (ppt) + K2SO4 net ionic: Cu 2+ + CrO4 2 CuCrO4 (ppt) the mass of precipitate will determine the original volume of the copper(ii) sulfate solution required 1mol 12.g 179.5g = moles ppt of CuCrO4 1K CrO 1CuSO 4 = moles of CuSO molCuSO M = L of CuSO4 solution, which is 107ml CuSO4 solution 8. molecular: CoCl2 + K2CrO4 CoCrO4 (ppt) + 2 KCl net ionic: Co 2+ + CrO4 2- CoCrO4 (ppt) the mass of precipitate will determine the original volume of the cobalt(ii) chloride solution.86g 1mol 175g = moles of CoCrO4 1K CrO 1Co = moles of K2CrO molK 2 0.5M = L of K2CrO4 solution, which is 6.0 ml K2CrO4 solution