Question

25 POINTS

Ammonia (NH3) reacts with oxygen (O2) to produce nitrogen monoxide (NO) and water (H2O). The balanced equation is 4NH3+5O2---> 4NO+6H2O.

How many liters of NO are produced when 2.0 liters of oxygen reacts with ammonia?

Please show all work

Answer 1

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

Step-by-step 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 (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 * (2.0\; \rm L) \cr &= \rm 1.6\; L\end{aligned}`.