A reactor is supplied with a feed of composition 46% C3H7O4, 44% O2 and 10% N2, if 50% of the limiting reagent is converted into product, what is the percentage of H2O in the product?

Correct Answer: 23
The reaction is 4C3H7O4 + 11O2 -> 12CO2 + 14H2O, => Limiting reagent is O2. Let 100 moles of feed was supplied, => after the reaction, moles of CO2 = 24, moles of H2O = 28, moles of O2 = 22, moles of C3H7O4 = 38, moles of N2 = 10, => percentage of H2O = 23%.