**Step 1. **mol Cl_{2} and mol F_{2}

**For Cl _{2}**

P = 337 mmHg = **0.443 atm**

${\mathbf{n}}_{{\mathbf{Cl}}_{\mathbf{2}}}\mathbf{=}\frac{\mathbf{PV}}{\mathbf{RT}}\phantom{\rule{0ex}{0ex}}{\mathbf{n}}_{{\mathbf{Cl}}_{\mathbf{2}}}\mathbf{=}\frac{(0.443{}\overline{)\mathrm{atm}})(2.05\overline{)L})}{(0.08206{\displaystyle \frac{\overline{)L}\xb7\overline{)\mathrm{atm}}}{\mathrm{mol}\xb7\overline{)K}}})(298\overline{)K})}$

Chlorine gas reacts with fluorine gas to form chlorine trifluoride.

Cl_{2}(g) + 3F_{2}(g) → 2ClF_{3}(g)

A 2.05 L reaction vessel, initially at 298 K, contains chlorine gas at a partial pressure of 337 mmHg and fluorine gas at a partial pressure of 801 mmHg.

Identify the limiting reactant and determine the theoretical yield of ClF_{3} in grams.

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