Answer
$$n_{Cl_2} = 3.959 \times 10^{-4} \space mol$$
Work Step by Step
- Calculate all the concentrations:
$$[CO] = ( 0.3500 )/(3.050) = 0.1148 M$$
$$[COCl_2] = ( 0.05500 )/(3.050) = 0.01803 M$$
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ COCl_2 ]}{[ CO ][ Cl_2 ]}$$
2. At equilibrium, these are the concentrations of each compound:
$ [ CO ] = 0.1148 \space M - x$
$ [ Cl_2 ] = - x$
$ [ COCl_2 ] = 0.01803 \space M + x$
$$1.2 \times 10^{3} = \frac{(0.01803 + x)}{(0.1148 - x)(- x)}$$
3. Solving for x:
$x_1 = -1.298 \times 10^{-4}$
$x_2 = 0.1158$
$x_2 = x$ would result in a negative concentration, thus: $x = x_1$
$$[Cl_2] = - (-1.298 \times 10^{-4}) = 1.298 \times 10^{-4} \space M$$
$$n_{Cl_2} = 1.298 \times 10^{-4} \space M \times 3.050 \space L = 3.959 \times 10^{-4} \space mol$$
