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)

Consider the following reaction:

2 SO_{2} (g) + O_{2}(g) → 2 SO_{3}(g)

282.8 mL of SO_{2} is allowed to react with 160.5 mL of O_{2} (both measured at 316 K and 50.1 mmHg ). If 185.0 mL of SO_{3} is collected (measured at 316 K and 50.1 mmHg ), what is the percent yield for the reaction?

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