KMnO4 and H2C2O4 react according to the following balanced equation: 2 KMnO4 + 5 H2C2O4 +...

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.

Solutions

Expert Solution

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

queen_honey_blossom answered 1 year ago

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