# How many mols of oxygen are required to react completely with 81 L of hydrogen to form water?

Approx. $40.5 \cdot L$ of dioxygen gas.
${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$
Given the stoichiometry, clearly there are $\text{2 equiv}$ dihydrogen gas per $\text{equiv}$ dioxygen gas.
Under the same conditions of temperature and pressure, $81 \cdot L$ of dihydrogen gas requires $40.5 \cdot L$ dioxygen gas for stoichiometric equivalence.