3NO₂(g) + H₂O(l) → 2HNO₃(l) + NO(g). If 3.9 mol NO₂ and 0.50 mol H₂O combine and react completely, how many moles of the excess reactant are present after the reaction is complete?

Apr 10, 2015

2.4 mol of ${\text{NO}}_{2}$ are present after the reaction is complete.

This is similar to an earlier question on Socratic, so I will simply go through the steps.

The balanced equation is

$3 \text{NO"_2+ "H"_2"O" → "2HNO"_3 + "NO}$

Step 1. Identify the limiting reactant.

$\text{From NO"_2: 3.9 cancel("mol NO₂") × "1 mol NO"/(3 cancel("mol NO₂")) = "1.3 mol NO}$

$\text{From H"_2"O": 0.50 cancel("mol H₂O") × ("1 mol NO")/(1cancel("mol H₂O")) = "0.50 mol NO}$

$\text{H"_2"O}$ is the limiting reactant, so ${\text{NO}}_{2}$ is the excess reactant.

Step 2. Calculate the moles of excess reactant used up.

${\text{Moles of NO"_2" used" = 0.50 cancel("mol H₂O") × ("3 mol NO"_2)/(1 cancel("mol H₂O")) = "1.50 mol NO}}_{2}$

Step 3. Calculate the moles of excess reactant left over.

$\text{Moles of excess NO"_2 = "3.9 mol – 1.50 mol" = "2.4 mol}$