Question

25 POINTS

Answer 1

Answer:

Assuming that O₂ is the limiting reagent, approximately 1.6 liters of NO will be produced.

Explanation:

Make sure that the equation has been balanced.

The coefficient in front of $\rm NO$ is $4$; the coefficient in front of $\rm O_2$ is $5$. The ratio between these two coefficients is equal to $\displaystyle \frac{4}{5} = 0.8$.

Assume that $\rm O_2$ is indeed the limiting reagent. In other words, assume that $\rm O_2$ runs out before all $\rm NH_3$ is consumed. The ratio between the number of moles of $\rm NO$ produced and the number of moles of $\rm O_2$ consumed will also be equal to $0.8$. That is:

$\displaystyle \frac{n(\mathrm{NO})}{n(\mathrm{O_2})} = \frac{\text{Coefficient of \mathrm{NO} }}{\text{Coefficient of \mathrm{O_2} }} = 0.8$.

Additionally, both $\rm NO$ and $\rm O_2$ are gases. Under the same temperature and pressure, one mole of each gas would occupy about the same volume. As a result,

$\displaystyle \frac{V(\mathrm{NO})}{V(\mathrm{O_2})} = \frac{n(\mathrm{NO})}{n(\mathrm{O_2})} = 0.8$.

The volume of $\rm O_2$ is already known. Hence, the volume of $\rm NO$ produced will be equal to:

$\begin{aligned} V(\mathrm{NO}) &= \frac{V(\mathrm{NO})}{V(\mathrm{O_2})} \cdot V(\mathrm{O_2}) \cr &= 0.8 \times (2.0\; \rm L) \cr &= \rm 1.6\; L\end{aligned}$.