Question

nitrogen monoxide is a pollutant commonly found in smokestack emissions. one way to remove it is to react it with ammonia.
4nh₃(g) + 6no(g) → 5n₂(g) + 6h₂o(ℓ)
how many liters of ammonia are required to change 28.0 l of nitrogen monoxide to nitrogen gas? assume 100% yield and that all gases are measured at the same temperature and pressure.
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Explanation:

Step1: Determine the mole ratio

From the balanced chemical equation \(4NH_3(g)+6NO(g)\to5N_2(g)+6H_2O(\ell)\), the mole ratio of \(NH_3\) to \(NO\) is \(4:6=\frac{2}{3}\).

Step2: Use the relationship for gases at same \(T\) and \(P\)

At the same temperature and pressure, \(V = n\times\frac{RT}{P}\), so \(V\propto n\) (Avogadro's law). Let \(V_{NH_3}\) be the volume of \(NH_3\) and \(V_{NO}\) be the volume of \(NO\). Then \(\frac{V_{NH_3}}{V_{NO}}=\frac{n_{NH_3}}{n_{NO}}\).
Given \(V_{NO} = 28.0\space L\), and \(\frac{n_{NH_3}}{n_{NO}}=\frac{4}{6}\), so \(V_{NH_3}=\frac{4}{6}\times V_{NO}\).

Step3: Calculate the volume of \(NH_3\)

Substitute \(V_{NO} = 28.0\space L\) into the formula: \(V_{NH_3}=\frac{4}{6}\times28.0\space L=\frac{4\times28.0}{6}\space L=\frac{112}{6}\space L\approx18.7\space L\)

Answer:

\(18.7\space L\)