# Problem: Consider the following reaction:2 SO2 (g) + O2(g) → 2 SO3(g)282.8 mL of SO2 is allowed to react with 160.5 mL of O2 (both measured at 316 K and 50.1 mmHg ). If 185.0 mL of SO3 is collected (measured at 316 K and 50.1 mmHg ), what is the percent yield for the reaction?

###### FREE Expert Solution

Recall the Ideal Gas Law equation:

$\overline{){\mathbf{PV}}{\mathbf{=}}{\mathbf{nRT}}}$

Step 1. actual yeild

$\mathbf{PV}\mathbf{=}\mathbf{nRT}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{PV}}{\mathbf{RT}}\mathbf{=}\frac{\mathbf{n}\overline{)\mathbf{RT}}}{\overline{)\mathbf{RT}}}\phantom{\rule{0ex}{0ex}}\overline{){\mathbf{n}}{\mathbf{=}}\frac{\mathbf{PV}}{\mathbf{RT}}}$

P = 50.1 mmHg = 0.066 atm
T = 316 K
R = 0.08206 (L·atm)/(mol·K)

###### Problem Details

Consider the following reaction:
2 SO2 (g) + O2(g) → 2 SO3(g)

282.8 mL of SO2 is allowed to react with 160.5 mL of O2 (both measured at 316 K and 50.1 mmHg ). If 185.0 mL of SO3 is collected (measured at 316 K and 50.1 mmHg ), what is the percent yield for the reaction?