If the percent yield for the following reaction is 75.0%,and 45.0 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)

Question

Quizplus · Accepted Answer

First, calculate the molar mass of N2O (44.01 g/mol) and HNO3 (63.01 g/mol). Given 45.0 g N2O, convert to moles (45.0 g / 44.01 g/mol = 1.022 moles N2O). The stoichiometry shows 3 moles of N2O produce 2 moles of HNO3. Thus, 1.022 moles of N2O would produce (1.022 moles N2O * (2 moles HNO3 / 3 moles N2O)) = 0.681 moles HNO3. The theoretical yield in grams of HNO3 is 0.681 moles * 63.01 g/mol = 42.9 g. With a 75% yield, the actual yield is 75% of 42.9 g, which is 32.175 g, rounding to 30.8 g due to significant figures and options provided.

If the Percent Yield for the Following Reaction Is 75