How many grams of oxygen are required to produce 55.6 g water vapor?

Jul 10, 2016

Approx. $50 \cdot g$ dioxygen are necessary.

Explanation:

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

$\text{Moles of water}$ $=$ $\frac{55.6 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ $3.09 \cdot m o l$

If there are $3.09 \cdot m o l$ of water, then clearly, there were $1.54 \cdot m o l$ ${O}_{2} \left(g\right)$ as a reactant.

And thus $\text{mass of oxygen}$ $=$ $1.54 \cdot m o l \times 32.00 \cdot g \cdot m o {l}^{-} 1$.