For the reaction PCl5(g) → PCl3(g) + Cl2(g)?
0.60 mole PCl5 is initially placed in a 1.00 L flask. At equilibrium, there is 0.16 mole PCl3 in the flask. The equilibrium concentrations of PCl5 and Cl2 are:
A. [Cl2] = 0.08 M; [PCl5] = 0.44 M
B. [Cl2] = 0.16 M; [PCl5] = 0.44 M
C. [Cl2] = 0.08 M; [PCl5] = 0.52 M
D. [Cl2] = 0.16 M; [PCl5] = 0.52 M
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Verified answer
PCl5 = (.6-.16)moles/1L = .44M
you had to make the same amount of Cl2 as you made PCl3
answer B
Since you formed 0.16 mol of PCl3, you had to also form 0.16 mol of Cl2. Also, you used up 0.16 mol of PCl5, so you have 0.44 mol of PCl5 remaining in the flask. So, B. is the best answer.