Question

KMnO4 and H2C2O4 react according to the following balanced equation: 2 KMnO4 + 5 H2C2O4 + 3 H2SO4 --> 2 MnSO4 + 10 CO2 + 8 H2O + K2SO4.

If 17.45 g of KMnO4 (MW = 158.03 g/mol) react with 28.85 g H2C2O4 (MW = 90.03 g/mol) how many grams of H2C2O4 will remain at the end of the reaction in the presence of excess H2SO4? Enter your answer to two decimal places.

Answer 1

moles of KMnO4 = 17.45 / 158.03 = 0.1104 mol

moles of H2C2O4 = 28.85 / 90.03 = 0.3204 mol

2 KMnO4 + 5 H2C2O4 + 3 H2SO4 -------------> 2 MnSO4 + 10 CO2 + 8 H2O + K2SO4.

      2                5                  3

0.1104        0.3204

moles of H2C2O4 remain = 0.3204 - (5 x 0.1104) / 2

                                         = 0.0444

mass of H2C2O4 remain = 0.0444 x 90.03

mass of H2C2O4 remain = 3.997 g