# How many grams of O_2 are required to produce 358.5 grams of ZnO in 2Zn + O_2 -> 2ZnO?

Mar 9, 2017

$Z n \left(s\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow Z n O \left(s\right)$
And we have $\frac{358.5 \cdot g}{81.41 \cdot g \cdot m o {l}^{-} 1} = 4.40 \cdot m o l$ of zinc oxide.
Given the stoichiometry, $2.20 \cdot m o l$ of dioxygen gas were necessary, i.e. $2.20 \cdot m o l \times 32.00 \cdot g \cdot m o {l}^{-} 1 = 70.0 \cdot g$