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Chlorine gas reacts with fluorine gas to form chlorine trifluoride. Cl2(g)+3F2(g)→2ClF3(g) A 1.60 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 877 mmHg . Part A Identify the limiting reactant and determine the theoretical yield of ClF3 in grams.

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Answer:

Limiting reagent: fluorine.

Theoretical yield of ClF₃: 4.62 g.

Explanation:

Hello,

In this case, one could identify the limiting reagent by comparing the available chlorine with the consumed chlorine by the fluorine considering the ideal gas equation and the 1 to 3 respectively stoichiometric relationship between them as shown below:

[tex]n_{Cl_2} ^{available}=\frac{337mmHg*\frac{1atm}{760mmHg}*1.60L}{0.082\frac{atm*L}{mol*K}*298K}=0.029molCl_2[/tex]

[tex]n_{Cl_2} ^{consumed \ by\ F_2 }=\frac{877mmHg*\frac{1atm}{760mmHg}*1.60L}{0.082\frac{atm*L}{molF_2*K}*298K}*\frac{1molCl_2}{3molF_2} =0.025molCl_2[/tex]

In such a way, there are 0.004 excess moles of chlorine, which means that fluorine is the limiting reagent. Therefore, the theoretical yield of ClF₃ turns out:

[tex]m_{ClF_3}=0.025molCl_2*\frac{2molClF_3}{1molCl_2}*\frac{92.448gClF_3}{1molClF_3}=4.62gClF_3[/tex]

Best regards.

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