Question

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.

Answer 1

Limiting reagent: fluorine.

Theoretical yield of ClF₃: 4.62 g.

Step-by-step 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:

`n_(Cl_2) ^(available)=(337mmHg*(1atm)/(760mmHg)*1.60L)/(0.082(atm*L)/(mol*K)*298K)=0.029molCl_2`

`n_(Cl_2) ^(consumed \ by\ F_2 )=(877mmHg*(1atm)/(760mmHg)*1.60L)/(0.082(atm*L)/(molF_2*K)*298K)*(1molCl_2)/(3molF_2) =0.025molCl_2`

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:

`m_(ClF_3)=0.025molCl_2*(2molClF_3)/(1molCl_2)*(92.448gClF_3)/(1molClF_3)=4.62gClF_3`

Best regards.