Question

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a sample of gas contains 0.1681 mol of h₂(g) and 0.1681 mol of o₂(g) and occupies a volume of 17.1 l. the following reaction takes place:
h₂(g) + o₂(g) → h₂o₂(g)
calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant and the reaction goes to the completion.
volume = l
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Explanation:

Step1: Write the balanced chemical equation

$$2H_{2}(g)+O_{2}(g)\to 2H_{2}O(g)$$

Step2: Determine the limiting reactant

The mole ratio of \(H_{2}\) to \(O_{2}\) in the reaction is \(2:1\). Given \(n_{H_{2}} = 0.1681\ mol\) and \(n_{O_{2}}=0.1681\ mol\). For \(H_{2}\), the amount of \(O_{2}\) required is \(n_{O_{2}\text{(required)}}=\frac{0.1681\ mol}{2}=0.08405\ mol\). Since \(0.08405\ mol<0.1681\ mol\) (available \(O_{2}\)), \(H_{2}\) is the limiting reactant.

Step3: Calculate moles of gas after reaction

Moles of \(H_{2}O\) formed: \(n_{H_{2}O}=0.1681\ mol\) (from \(H_{2}\) as limiting reactant). Moles of \(O_{2}\) remaining: \(n_{O_{2}\text{(remaining)}}=0.1681 - 0.08405=0.08405\ mol\). Total moles of gas after reaction: \(n_{total}=0.1681 + 0.08405=0.25215\ mol\).

Step4: Use Avogadro's law (\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\))

Initial moles \(n_{1}=0.1681 + 0.1681=0.3362\ mol\), \(V_{1} = 17.1\ L\), \(n_{2}=0.25215\ mol\). Then \(V_{2}=\frac{n_{2}V_{1}}{n_{1}}=\frac{0.25215\ mol\times17.1\ L}{0.3362\ mol}\)
$$V_{2}=\frac{0.25215\times17.1}{0.3362}\ L$$
$$V_{2}=12.8\ L$$

Answer:

\(12.8\)