Nitrogen dioxidegasand liquid water react to form aqueous nitric acid and nitrogen monoxidegas. Suppose you have of and of in a reactor. Calculate the largest amount of that could be produced. Round your answer to the nearest .

3.3 mol

Explanation:

5.0 mol of NO2

7.0 mol of H2O

The balanced equation is given as;

3NO2 + H2O → 2HNO3 + NO

From the reaction;

3 mol of NO2 requires 1 mol of H2O to  produce 2 mol of HNO3

If all the NO2 were used up, there would still be an excess of H2O, so the limiting reactant is NO2.

if 3 mol  NO2 produces 2 mol H2O

5 mol  NO2 would produce x mol

x = (5 * 2) / 3 = 3.33 mol

To the nearest 0.1 mol, the answer is 3.3 mol