# In the combustion of hydrogen gas, hydrogen reacts with oxygen from the air to form water vapor. Hydrogen + Oxygen -> water. If you burn 58.1 g of hydrogen and produce 519 g of water how much oxygen reacted?

Jul 7, 2016

${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$

Clearly, approx. $500$ $g$ of dioxygen reacted.

#### Explanation:

The stoichiometrically balanced equation tells us unequivocally that $1$ $m o l$ dihydrogen reacts with $0.50$ $m o l$ dioxygen to give $1$ $m o l$ water.

$\text{Moles of dihydrogen}$ $=$ $\frac{58.1 \cdot g}{2.02 \cdot g \cdot m o l}$ $=$ $28.7 \cdot m o l$ ${H}_{2}$

$\text{Moles of water}$ $=$ $\frac{519.0 \cdot g}{18.02 \cdot g \cdot m o l}$ $=$ $28.7 \cdot m o l$ ${H}_{2} O$

The water is stoichiometric with respect to the amount of dihydrogen. (What do I mean by this?) There was thus a stoichiometric quantity of dioxygen gas, i.e. $14.4 \cdot m o l$ ${O}_{2}$

$\text{Mass of dioxygen}$ $=$ $14.4 \cdot \cancel{m o l} \times 32.00 \cdot g \cdot \cancel{m o {l}^{-} 1}$ $=$ ??g