Question

In a container 2L of $N_2$ and 6L of $H_2$ are allowed to react to form $NH_3$ gas. what is the mass of $NH_3$ Produced at STP?

Answer 1

$\frac{85}{28}$ g

Answer 2

The balanced chemical equation for the formation of ammonia is $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$. According to Avogadro's Law, at constant temperature and pressure, the volume of gases is directly proportional to the number of moles. Thus, the volume ratio of reacting gases is equal to their mole ratio. The stoichiometric volume ratio is $V(N_2) : V(H_2) : V(NH_3) = 1 : 3 : 2$. Given volumes are $V(N_2) = 2$ L and $V(H_2) = 6$ L. The ratio of given volumes is $2:6$, which simplifies to $1:3$. This matches the stoichiometric ratio, meaning both reactants are present in stoichiometric amounts and will be completely consumed. From the stoichiometry, 1 volume of $N_2$ produces 2 volumes of $NH_3$. Therefore, 2 L of $N_2$ will produce $2 \times 2 = 4$ L of $NH_3$. At Standard Temperature and Pressure (STP), the molar volume of an ideal gas is 22.4 L/mol. The number of moles of $NH_3$ produced is $\frac{\text{Volume of } NH_3}{\text{Molar volume at STP}} = \frac{4 \text{ L}}{22.4 \text{ L/mol}}$. The molar mass of $NH_3$ is $14 (\text{for } N) + 3 \times 1 (\text{for } H) = 17$ g/mol. The mass of $NH_3$ produced is: $\text{Mass of } NH_3 = \text{Moles of } NH_3 \times \text{Molar mass of } NH_3$ $\text{Mass of } NH_3 = \left(\frac{4}{22.4}\right) \text{ mol} \times 17 \text{ g/mol}$ $\text{Mass of } NH_3 = \frac{40}{224} \times 17 \text{ g}$ $\text{Mass of } NH_3 = \frac{5}{28} \times 17 \text{ g}$ $\text{Mass of } NH_3 = \frac{85}{28} \text{ g}$