Let's analyze the given chemical reaction and understand what happens when [tex]\( H_2O \)[/tex] is removed from it.
The balanced chemical reaction is:
[tex]\[ 2 H_2 + O_2 \rightleftharpoons 2 H_2O \][/tex]
This reaction is at equilibrium, which means the rate of the forward reaction (producing [tex]\( H_2O \)[/tex] from [tex]\( H_2 \)[/tex] and [tex]\( O_2 \)[/tex]) is equal to the rate of the reverse reaction (breaking down [tex]\( H_2O \)[/tex] into [tex]\( H_2 \)[/tex] and [tex]\( O_2 \)[/tex]).
According to Le Chatelier's principle, if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
If we remove [tex]\( H_2O \)[/tex] from the reaction mixture, the equilibrium will be disturbed because we are decreasing the concentration of one of the products.
To counteract this change, the system will shift the equilibrium to the right in order to produce more [tex]\( H_2O \)[/tex]. This means that the forward reaction (where [tex]\( H_2 \)[/tex] and [tex]\( O_2 \)[/tex] react to form [tex]\( H_2O \)[/tex]) will be favored.
Hence, the correct answer is:
A. The reactants will react to produce more [tex]\( H_2O \)[/tex].
