Question

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a sample of gas contains 0.1534 mol of n₂(g) and 0.4602 mol of h₂(g) and occupies a volume of 23.1 l. the following reaction takes place:
n₂(g) + 3h₂(g) → 2nh₃(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: Determine the limiting reactant

From the balanced equation \(N_2(g)+3H_2(g)\to2NH_3(g)\), the mole ratio of \(N_2\) to \(H_2\) is \(1:3\).
Given \(n_{N_2}=0.1534\ mol\) and \(n_{H_2}=0.4602\ mol\).
If \(N_2\) is the limiting reactant, \(n_{H_2}\) required \(=3\times0.1534 = 0.4602\ mol\).
Since the available \(n_{H_2}=0.4602\ mol\), both reactants are completely consumed.

Step2: Calculate the moles of \(NH_3\) produced

From the balanced equation, \(1\ mol\ N_2\) produces \(2\ mol\ NH_3\).
Since \(n_{N_2}=0.1534\ mol\), \(n_{NH_3}=2\times0.1534 = 0.3068\ mol\).

Step3: Use Avogadro's law (\(V = n\times V_m\), assuming \(V_m\) is constant as temperature and pressure are constant)

Initially, total moles \(n_{initial}=n_{N_2}+n_{H_2}=0.1534 + 0.4602=0.6136\ mol\), \(V_{initial}=23.1\ L\).
Finally, \(n_{final}=n_{NH_3}=0.3068\ mol\).
Using \(\frac{V_{initial}}{n_{initial}}=\frac{V_{final}}{n_{final}}\), \(V_{final}=\frac{n_{final}}{n_{initial}}\times V_{initial}\).
\(V_{final}=\frac{0.3068}{0.6136}\times23.1\)
\(V_{final}= 11.55\ L\)

Answer:

\(11.55\)