## General Chemistry: Principles and Modern Applications (10th Edition)

$n_{SbCl_3} = 0.164 \space mol$ $n_{Cl_2} = 0.0443 \space mol$ $n_{SbCl_5} = 0.116 \space mol$
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: $[ SbCl_3 ] = 0.112 - 0.0463 = 0.0657 \space M$ $[ Cl_2 ] = 0.0640 - 0.0463 = 0.0177 \space M$ $[ SbCl_5 ] = 0.0463 \space M$ $n_{SbCl_3} = 2.50 \times 0.0657 = 0.164 \space mol$ $n_{Cl_2} = 2.50 \times 0.0177 = 0.0443 \space mol$ $n_{SbCl_5} = 2.50 \times 0.0463 = 0.116 \space mol$