Answer
1.00 g of $CO_2$ at equilibrium.
Work Step by Step
1. Calculate each molarity:
$ CO $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 28.01 g/mol
- Calculate the amount of moles:
$$ 1.00 \space g \times \frac{1 \space mol}{ 28.01 \space g} = 0.0357 \space mol$$
- Calculate the molarity:
$$ \frac{ 0.0357 \space mol}{ 1.41 \space L} = 0.0253 \space M $$
$ H_2O $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol
- Calculate the amount of moles:
$$ 1.00 \space g \times \frac{1 \space mol}{ 18.02 \space g} = 0.0555 \space mol$$
- Calculate the molarity:
$$ \frac{ 0.0555 \space mol}{ 1.41 \space L} = 0.0394 \space M $$
$ H_2 $ : ( 1.008 $\times$ 2 )= 2.016 g/mol
- Calculate the amount of moles:
$$ 1.00 \space g \times \frac{1 \space mol}{ 2.016 \space g} = 0.496 \space mol$$
- Calculate the molarity:
$$ \frac{ 0.496 \space mol}{ 1.41 \space L} = 0.352 \space M $$
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ CO_2 ][ H_2 ]}{[ CO ][ H_2O ]}$$
2. At equilibrium, these are the concentrations of each compound:
$ [ CO ] = 0.0253 \space M - x$
$ [ H_2O ] = 0.0555 \space M - x$
$ [ CO_2 ] = 0 \space M + x$
$ [ H_2 ] = 0.496 \space M + x$
$$23.2 = \frac{(x)(0.496 + x)}{(0.0253 - x)(0.0555 - x)}$$
3. Solve for x:
$x_1 = 0.0162$
$x_2 = 0.0906$
x cannot be greater than $0.0253$, because that would result in a negative concentration.
$x = x_1 = 0.0162$
4. Find the mass of $CO_2$:
$ [ CO_2 ] = x = 0.0162 \space M$
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
- Use all the information as conversion factors:
$$ 1.41 \space L \times \frac{ 0.0162 \space mol}{1 \space L} \times \frac{ 44.01 \space g}{1 \space mol} = 1.00 \space g$$
