2 SO2(g) + O2(g) >< 2 SO3(g)

Question

In order to obtain the equilibrium system above, 2.60 mol of SO2(g) and 2.30 mol of O2(g) are injected into a 1.00 L container. When the system reaches equilibrium, the concentration of the remaining SO2(g) is 1.41 mol/L. The concentration of O2(g) at equilibrium is _.__ mol/L

need help solving this problem. Help much appreciated

Answer 1

2SO₂(g) + O₂(g) <===> 2SO₃(g)

2.6.............2.3.....................0.........Initial

-2x............-x.......................+2x......Change

1.41...........2.3-x...................................Equilibrium

Since we are told the initial and final [SO2], we can easily determine how much was used up.

2.6 - 2x = 1.41

2x = 1.19

x = 0.595

Since the equilibrium [O₂] is 2.3 - x, it will be 2.3 - 0.595

[O₂] = 1.71 mol/L