Question

0. If 5.6 kg or Account 10. If 5.6 kg of nitrogen sulfate reacts with 1.3 kg of potassium permanganate, determine the products. (Hint: there's an extra conversion step!) a. Find the limiting reagent. (2 pts) b. Calculate the amount of permanganate-containing product is produced. (2 pts) c. Calculate the amount of excess. (2 pts) d. If 0.56 kg of permanganate-containing product is actually produced, calculate the percent yield (2 pts).

Answer 1

Write the reaction between aluminum sulfate and potassium permanganate as follows: A1,(S0.)+ 6KMnO4 - 2A1(MnO4)3 +3K,SO. Mass of A1, (SO4)3 = 5.6 Kg molar mass of A1, (SO4) = 342.15 g/mol Mass of A1, (SO) moles of Al(SO4)3 = molar mass of A1, (SO) 5.6 Kg , 1000g 342.15 g/mol 1 Kg = 16.4 mol Mass of KMnO4 = 1.3 Kg molar mass of KMnO4 = 158.0 g/mol Mass of KMnO4 moles of KMnO4 = molar mass of KMnO4 1.3 Kg 1000 g 158.0 g/mol Kg = 8.23 mol (a) From the reaction the stoichiometric relation between Al2(SO4), and KMnO, is moles of Al, (S04), moles of KMnO. Substitute the calculated moles in the above relation, the moles with least is the limitig reagent and the moles with highest is the excess reagent. 16.4 mol A1, (SO.) 8.23 mol KMnO, 1 16.4 mol A1, (50.), > 1.37 mol KMnO4 Therefore, the limiting reactant is KMnO4! (b) From the reaction the stoichiometric relation between Al (MnO4), and KMnO, is moles of Al( MO), moles of KMnO. moles of A1 (MnO.), = 8:23 = 2.74mol molar mass of A1(MnO4), = 383.78 g/mol mass of Al MnO4) = 383.78 g/mol x 2.74mol = 105 g = 1.05 Kg Therefore, the amount of A1(MnO.), produced is 1.05 Kg

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Unreacted moles of A1,(SO.)= moles of A12(SO4) -moles of A1, (S0.), consumed = 16.4 mol -1.37 mol = 15.03mol molar mass of A1, (S04), = 342.15 g/mol mass of A1, (SOx), remained = 342.15 g/mol *15.03 mol = 5142g = 5.142 Kg Therefore, the requried solution is 5. 142Kg A1, (S04), in excess. Given actual yield= 0.56 Kg theoretical yield = 1.05 Kg Calculate % yield as follows: actual yield %yield = theoretical yield -X100% 0.56 Kg x100% %yield = 1 1.05 Kg %yield = 53.3% therefore, the required solution is 53.3%