If the percent yield for the following reaction is 75.0%, and 40.0 g of NO2 are consumed in the reaction, how many grams of nitric acid, HNO3(aq) are produced? 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) If the percent yield for the following reaction is 75.0%, and 40.0 g of NO2 are consumed in the reaction, how many grams of nitric acid, HNO3(aq) are produced? 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g)

Explanation:

Hello,

In this case, for the chemical reaction:

The first step is to compute the theoretical yield of nitric acid via stoichiometry in terms of the 3:2 ratio between nitrogen dioxide (molar mass = 46 g/mol) and nitric acid (molar mass = 63 g/mol) respectively:

Now, the actual amount is computed by taking into account the 75.0-% percent yield:

Best regards.

the product of the pressure and volume is also constant