Question

part a
a 24.0 g sample of nitrogen gas reacts with an excess of hydrogen gas to give an actual yield of 3.85 g nh₃. what is the percent yield for this reaction?
reaction: n₂(g) + 3 h₂(g) → 2 nh₃(g)
26.4%
73.6%
13.2%
86.8%
none of the above
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Explanation:

Step1: Calculate Moles of \(N_2\)

Molar mass of \(N_2 = 2\times14.01 = 28.02\space g/mol\).
Moles of \(N_2=\frac{24.0\space g}{28.02\space g/mol}\approx0.8565\space mol\).

Step2: Determine Theoretical Yield of \(NH_3\)

From reaction \(N_2 + 3H_2
ightarrow2NH_3\), 1 mol \(N_2\) produces 2 mol \(NH_3\).
Moles of \(NH_3\) (theoretical) \(= 2\times0.8565 = 1.713\space mol\).
Molar mass of \(NH_3 = 14.01 + 3\times1.008 = 17.034\space g/mol\).
Theoretical yield \(= 1.713\space mol\times17.034\space g/mol\approx29.18\space g\).

Step3: Calculate Percent Yield

Percent yield \(=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100=\frac{3.85\space g}{29.18\space g}\times100\approx13.2\%\).

Answer:

13.2% (Option: 13.2%)