Question

LAB VIL PRACTICE PROBLEMS NAME: SECTION: PART D. CHAPTER 4 PROBLEMS 1) Reaction Stoichiometry, Limiting Reactant, Theoretical and Percent Yields For the equation Cu + C > Cu + CO, if 2.73 moles of Care reacted with an excess of Cuzo, how many moles of CO would be produced? moles b) For the equation Co(NO) + (NH),S COS; + NH.NO, if 79.2 of CNO are reacted with an excess of (NH),S, how many moles of NH.NO, would be produced moles el For the equation (NH)CO, CO + H2O + NH, in 4.01 moles of (NH.CO reacted, what is the theoretical yield of CO,? d) For the equation N H, NH, + Nx, if the theoretical yield is 28.0 grams of Nwhat mass in grams of N H, would have to react? e) For the equation FeS + HCI → FeCl2 + H,S, if 35.0g of FeS is added to 350g of HCl and allowed to react, what is the theoretical yield of H2S? Using the numbers from problem e) above, if 10.2 g of H,S were obtained in the lab (actual yield), what is the percent yield? 8) How many grams of K C 04. potassium oxalate, would contain 182 grams of pure potassium? II) Solution Concentration and Solution Stoichiometry a) What is the molarity of a solution made by dissolving 6,08g of sodium acetate, CH,COONa, in water, and diluting it to a total volume of 0.750L? _

Answer 1

1. C4.0 + c I mole of a produce = 2.73 moles of a produce 204 + co I mole of 10 2.73 moles of co Hence, 2.73 moles of c will produce. 2CO (NO3)3 +3 (NH4)2S CORS3 + 6N Hq NO 3 molar man of co (NO3)3 = neraugonal 244,96 g/mol mous of CO(NO3)3 = 79.29 moles of COCNO 3/3 = mas = 19.29 = 0.3233 mol molar man 244.96 glmol. 2 moles of COCNO313 produce 6 moles of NHUNO3 = 0.3233 moles of CO (NO3)3 produce moles of NH Y NO3 = = 6 mol NHU NO3 x 0.323 mol co(NO3)3 2 mol co (NO3/3 - 0.9699 mol = 0.97 mol NHU NO 3 Answer (NH4)2CO3 CO + H₂O + 2NH3 molar man of (02 = 44 gimol. I mole of (NH4)2CO3 produce I mole of co2 I mole of (NH4)₂CO3 produce 44 gof co2

4.0lmol of (NH4)2CO3 produce man of CO₂ = = 44 g CO2 X 4,01 mol (NH4)2CO3 Imol (NH4)2CO3 = 176.44 g CO2 Hence theoretical yreld of CO2 is 176.449 d) 3 Na Hy 4NH3 + N2 molar man of N₂ = 28 g/mol molar mans of NaHy = 32 g/mol 1 moles of No produced by 3 moles of Na Hy 28 g of No produced by 3x32g of Naty = man of Nå Hy that react = 3X32 g = 969 Natu