What volume of O_2 at 912 mmHg and 21° C is required to synthesize 18.0 mol of NO?

May 16, 2016

${V}_{{O}_{2}} = 181 L$

Explanation:

Assuming the nitrogen monoxide is made from oxygen and nitrogen according to the following reaction:

${N}_{2} \left(g\right) + {O}_{2} \left(g\right) \to 2 N O \left(g\right)$

Therefore, 2 moles of $N O$ are prepared using only 1 mole of ${O}_{2}$, thus, we will need 9.00 moles of ${O}_{2}$ to prepare 18.0 moles of $N O$.

The pressure of ${O}_{2}$ is P=912cancel(mmHg)xx(1atm)/(760cancel(mmHg)=1.20atm

The Kelvin temperature is $T = 21 + 273 = 294 K$

Therefore, using the ideal gas law, $P V = n R T$, the volume of oxygen can be found by:

$V = \frac{n R T}{P} = \frac{9.00 \cancel{m o l} \times 0.0821 \frac{\cancel{a t m} \cdot L}{\cancel{m o l} \cdot \cancel{K}} \times 294 \cancel{K}}{1.2 \cancel{a t m}} = 181 L$