First, write the balanced chemical equation: $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$. Calculate the number of moles: Moles of $H_2 = 10 \text{ g} / 2 \text{ g/mol} = 5 \text{ mol}$; Moles of $O_2 = 64 \text{ g} / 32 \text{ g/mol} = 2 \text{ mol}$. According to the stoichiometry, 1 mole of $O_2$ reacts with 2 moles of $H_2$. Therefore, 2 moles of $O_2$ require 4 moles of $H_2$. Since we have 5 moles of $H_2$, $O_2$ is the limiting reagent and $H_2$ is in excess. The amount of product formed depends on the limiting reagent ($O_2$). From the equation, 1 mole of $O_2$ gives 2 moles of $H_2O$. Thus, 2 moles of $O_2$ will produce 4 moles of $H_2O$. Mass of $H_2O = 4 \text{ mol} \times 18 \text{ g/mol} = 72 \text{ g}$.