Question

How many grams of nitric acid can be prepared from the reaction of 138 grams of NO2 with 54 grams of H2O according to the equation: 3NO2 + H2O -> 2HNO3 + NO

Answer 1

n(NO2) = 138/46 = 3 mol

n(H2O) = 54/18 = 3 mol

water in excess, therefore:

n(HNO3) = n(NO2)*2/3 = 2 mol

m(HNO3) = 2*63 = 126 g