Answer
$n_{SbCl_3} = 0.164 \space mol$
$n_{Cl_2} = 0.0443 \space mol$
$n_{SbCl_5} = 0.116 \space mol$
Work Step by Step
1. Calculate all the concentrations:
$$[SbCl_3] = ( 0.280 )/(2.50) = 0.112 M$$
$$[Cl_2] = ( 0.160 )/(2.50) = 0.0640 M$$
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ SbCl_3 ][ Cl_2 ]}{[ SbCl_5 ]}$$
2. At equilibrium, these are the concentrations of each compound:
$ [ SbCl_3 ] = 0.112 \space M - x$
$ [ Cl_2 ] = 0.0640 \space M - x$
$ [ SbCl_5 ] = 0 + x$
$$2.5 \times 10^{-2} = \frac{(0.112 - x)(0.0640 - x)}{x}$$
3. Solve for x:
$x_ 1= 0.0463$
$x_2 = 0.155$
x cannot be greater than 0.0640, therefore, $x = x_1$
4. Substitute and find the equilibrium amounts:
$ [ SbCl_3 ] = 0.112 - 0.0463 = 0.0657 \space M$
$ [ Cl_2 ] = 0.0640 - 0.0463 = 0.0177 \space M$
$ [ SbCl_5 ] = 0.0463 \space M$
$n_{SbCl_3} = 2.50 \times 0.0657 = 0.164 \space mol$
$n_{Cl_2} = 2.50 \times 0.0177 = 0.0443 \space mol$
$n_{SbCl_5} = 2.50 \times 0.0463 = 0.116 \space mol$
