Question

what quantity, in moles, of hno₃ will be produced from the reaction of 18.5 g of no₂ with excess water in the following chemical reaction? 3no₂(g) + h₂o(l) ⇌ 2hno₃(g) + no(g)

Explanation:

Step1: Calculate moles of \(NO_2\)

The molar mass of \(NO_2\) is \(M=(14 + 2\times16)\space g/mol=46\space g/mol\).
Using the formula \(n=\frac{m}{M}\), where \(m = 18.5\space g\) and \(M = 46\space g/mol\), we have \(n_{NO_2}=\frac{18.5\space g}{46\space g/mol}\approx0.402\space mol\).

Step2: Use mole - ratio from the balanced equation

From the balanced equation \(3NO_2(g)+H_2O(l)
ightleftharpoons2HNO_3(g)+NO(g)\), the mole ratio of \(NO_2\) to \(HNO_3\) is \(3:2\).
Let \(n_{HNO_3}\) be the moles of \(HNO_3\). Then \(\frac{n_{HNO_3}}{n_{NO_2}}=\frac{2}{3}\).
Substituting \(n_{NO_2} = 0.402\space mol\) into the equation, we get \(n_{HNO_3}=\frac{2}{3}\times n_{NO_2}\).
\(n_{HNO_3}=\frac{2}{3}\times0.402\space mol\approx0.268\space mol\).

Answer:

\(0.268\space mol\)