Answer
(a) Since $Q_c \lt K_c$, the mixture is not at equilibrium;
(b) And the net change will occur in the products direction.
Work Step by Step
1. Let's call the equal mass 'm', then the number of moles of each compound is:
$$n_{compound}= \frac{m}{M(compound)}$$
2. Calculate each molar mass:
$ CO $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 28.01 g/mol
$ H_2O $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
$ H_2 $ : ( 1.008 $\times$ 2 )= 2.016 g/mol
3. Write the Q expression for this reaction, where the equilibrium does not depend on the volume:
$$Q = \frac{n_{CO_2}n_{H_2}}{n_{CO}n_{H_2O}} = \frac{ \frac{m}{44.01} \frac{m}{2.016} }{ \frac{m}{28.01} \frac{m}{18.02}} = \frac{ \frac{1}{44.01} \frac{1}{2.016} }{ \frac{1}{28.01} \frac{1}{18.02}} = 5.69$$
Since $Q_c \lt K_c$, the mixture is not at equilibrium, and the net change will occur in the products direction, in order to increase $Q_c$.
