Equilibrium Question

Question

For the reaction

PCl3​(g)+Cl2​(g)PCl5​(g)

the initial mixture contains 5.90 mol PCl₃, 5.53 mol Cl₂, and no moles of PCl₅. At equilibrium, 1.38 mol of PCl₃ remains.

 How many moles of Cl₂ and PCl₅ are present at equilibrium? (Be sure to give your answer to two decimal places.)

 ____ mol Cl₂   

 ____ mol PCl₅   

Answer 1

Start off by putting together an equilibrium table. First row being the initial number of moles (Since same volume). Second row being the change (in terms of x). Note the change is negative for reactants and positive for products and that x is multiplied by the stoichiometric coefficient. Finally, by summing the first and second rows, you will get the number of moles at equilibrium in terms of x.

PCl3​(g)+Cl2​(g)PCl5​(g)

5.9 mol 5.53 mol 0

-x -x x

5.9-x 5.53-x x

Now using the information that PCl3 is 1.38 moles at equilibrium, you can equate and solve for x.

5.9-x=1.38 => x=4.52 mol

Then nCl2=5.53-x=5.53-4.52= 1.01 mol

nPCl5=x=4.52 mol