Discussion
2.8 g of N2(g) and 0.8 g of H2(g) are allowed to react completely. What is the mass of NH3(g) formed? (molar mass of N2(g)=28, H2(g)=2 and NH3(g)=17)
Answer: B
The mass of ammonia formed is 3.4 g.
\(\text{N}_{2} + 3\text{H}_{2} \rightarrow 2\text{NH}_{3}\). The moles are \(\frac{2.8}{28} = 0.1\) of nitrogen and \(\frac{0.8}{2} = 0.4\) of hydrogen.
0.1 mol of nitrogen needs only 0.3 mol of hydrogen, so nitrogen is the limiting reagent.
\[ \begin{aligned} n(\text{NH}_{3}) &= 2 \times 0.1 = 0.2\ \text{mol} \\ m &= 0.2 \times 17 = 3.4\ \text{g} \end{aligned} \]
3.6 g would be the total mass of the reactants, which is more than can react because 0.1 mol of hydrogen is left over.
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