What mass, in grams, of NO2 is needed to form 180.4 g of NO? Assume excess water. 3NO2(g) + H2O(l) → HNO3(aq) + NO(g)

Answer 1

The first step to answering this question is to write out a balanced chemical equation.

3NO₂(g) + H₂O(l) > 2HNO₃(aq) + NO(g)

To assess how much NO₂ is needed, we need to know the moles of NO formed. So we convert the grams of NO into the number of moles using the molecular weight of NO (= 30.01 g/mol)

Thus, 180.4 g NO × (1 mol NO/30.01 g NO) = 180.4/30.01 mol NO = 6.011 mol NO

Using the coefficients of the balanced equation we see that for every 3 moles of NO₂(g) consumed in the reaction, 1 mol of NO(g) is formed. Thus, mol NO₂ = 3 × mol NO = 3 × 6.011 = 18.033 mol NO₂

Then, convert the moles of NO₂ back into grams using the molecular weight (= 46.006 g/mol)

18.033 mole NO₂ × 46.006 g NO₂/1 mol NO₂ = 18.033 × 46.006 = 829.626 g NO₂