Question

nitrogen dioxide (no₂) gas and liquid water (h₂o) react to form aqueous nitric acid (hno₃) and nitrogen monoxide (no) gas. suppose you have 5.0 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.

Explanation:

Step1: Write the balanced chemical equation

$$3NO_{2}+H_{2}O = 2HNO_{3}+NO$$

Step2: Determine the limiting reactant

For \(NO_{2}\):
If all \(5.0\space mol\) of \(NO_{2}\) reacts, the moles of \(HNO_{3}\) produced is \(n_{HNO_{3}}=\frac{2}{3}\times5.0\space mol\approx3.3\space mol\)
For \(H_{2}O\):
If all \(13.0\space mol\) of \(H_{2}O\) reacts, the moles of \(HNO_{3}\) produced is \(n_{HNO_{3}} = 2\times13.0\space mol = 26.0\space mol\)
Since \(NO_{2}\) gives a smaller amount of \(HNO_{3}\), \(NO_{2}\) is the limiting reactant.

Answer:

\(3.3\space mol\)