Question

6. Balance the following equation: Sc(NO) Na2SO4 NaNO + Se2(SO4)s + The following information may be helpful for problems 5-6. Molar mass of Na2S203= 158.11 g/mol. Molar mass of AgBr 187.78 g/mol. Molar mass of Na3Ag(S2O3)2 = 401.10 g/mol. Molar mass of NaBr 102.90 g/mol. 7. Consider the following balanced equation: 2NA2S2O3 AgBr Na3Ag(S203)2 NaBr + + What is the limiting reagent or reactant if 52.36 grams of Na2S2O3 reacts with 29.47 grams of AgBr to form 6.15 g of NaBr? to 2om S0.0 bnuot 8. What is the percent yield if 52.36 grams of Na2 S2 O3 reacts with 29.47 grams of AgBr to form 6.15 g of NaBr? Pels

Answer 1

6. Balance the following equation: 2 Sc(NO), + 3 Na2SO4 → Sc2(SO4), +6 NaNO The following information may be helpful for problems 5-6. Molar mass of Na2S2O3 = 158.11 g/mol. Molar mass of AgBr = 187.78 g/mol. Molar mass of NazAg(S203)2 =401.10 g/mol. Molar mass of NaBr = 102.90 g/mol. 7. Consider the following balanced equation: 2Na2S203 + AgBr - Na3Ag(S203)2 + NaBr What is the limiting reagent or reactant if 52.36 grams of Na2S2O3 reacts with 29.47 grams of AgBr to form 6.15 g of NaBr? From the balanced equation, 2 mole of Na2s203 = 1 mole of AgBr i.e 316.22 gms of Na2S203 = 187.78 gms of AGBP 52.36 gms of Na2S203 = 52.36*187.78/316.22 = 31.09 gms of AgBr. Therefore, AgBr is the limiting reagent and the amount of the NaBr formed depends on limiting reagent only. The amount of NaBr to formed = 102.90* 29.47/187.78 = 16.15 gms 8. What is the percent yield if 52.36 grams of Na S03 reacts with 29.47 grams of AgBr to form 6.15 g of NaBr? Percet yield = 6.15*100/16.15 = 38.08 %