Describe the effect of:
$(a)$ Addition of $H_2$
$(b)$ Addition of $CH_3OH$
$(c)$ Removal of $CO$
$(d)$ Removal of $CH_3OH$ on the equilibrium of the reaction:
$2H_{2(g)} + CO_{(g)} \longleftrightarrow CH_3OH_{(g)}$

(N/A) According to Le Chatelier's principle,on addition of $H_2$,the concentration of reactants increases,so the equilibrium will shift in the forward direction to consume the added $H_2$.
$(b)$ On addition of $CH_3OH$,the concentration of the product increases,so the equilibrium will shift in the backward direction to counteract the increase.
$(c)$ On removing $CO$,the concentration of a reactant decreases,so the equilibrium will shift in the backward direction to produce more $CO$.
$(d)$ On removing $CH_3OH$,the concentration of the product decreases,so the equilibrium will shift in the forward direction to produce more $CH_3OH$.