solving moles to moles limiting reactant problems

Question

Nitrogen dioxide (NO2) gas and liquid water (H2O) react to form aqueous nitric acid (HNO3) and nitrogen monoxide (NO) gas. Suppose you have 11.0 mol of NO2 and 13.0 mol of H2O in a reactor.

Calculate the largest amount of HNO3 that could be produced. Round your answer to the nearest 0.1 mol.

Answer 1

1). Write a correctly balanced equation:

3NO₂(g) + H₂O(l) ==> 2HNO₃(aq) + NO(g) ... balanced equation

2). Find limiting reactant. Easy way is to divide mols of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less is the limiting reactant.

NO₂: 11.0 mols (÷3->3.7)

H₂O: 13.0 mols (÷1->13)

NO₂ is limiting

3). Use the moles of the limiting reactant to find the amount of product that can form.

11.0 mols NO₂ x 2 mols HNO₃ / 3 mols NO₂ = 7.3 mols HNO₃