Question

3K2MnO4 + 4CO2 + 2H2O --> 2KMnO4 +4KHCO3 + MnO2

A.) Which reactant would be the limiting reactant if 9.50 moles of K2MnO4, 7.02 x 10^24 molecules of CO2, and 60.0g of H2O were chemically combined?

B.) If the percent yield for MnO2 is 69.2% for the above reaction, how many grams of MnO2 are actually produced?

Answer 1

Number of moles of K2MnO4 = 9.50 moles

Number of moles of CO2 = 7.02 * 10^(24)/6.023 * 10^(23) = 11.65 moles

Number of moles of H2O = 60/18 = 3.333 moles

H2O is the limiting reagent in the reaction

B) 2 moles of water will give 1 mole of MnO2

Moles of MnO2 formed = 3.333/2 = 1.6666 moles (Yield - 100%)

Moles of MNO2 formed (Actual Yield) = 1.6666 * 69.2/1000 = 1.1533 moles

Molar mass of MnO2 = 86.93 gm/mol

Mass of MnO2 produced = 1.1533 moles * 86.93 gm/mol = 100.259 grams