# 8934*g of oxygen gas is reacted with stoichiometric zinc sulfide. What mass of zinc oxide results?

May 10, 2017

We interrogate the reaction............and get a mass of over $18 \cdot k g$ with respect to $\text{zinc oxide}$.

#### Explanation:

$2 Z n S \left(s\right) + 3 {O}_{2} \left(g\right) \rightarrow 2 Z n O \left(s\right) + 2 S {O}_{2} \left(g\right)$

$\text{Moles of dioxygen} = \frac{8934 \cdot g}{32.00 \cdot g \cdot m o {l}^{-} 1} = 279.2 \cdot m o l$.

And thus we represent an equivalent quantity of $\text{zinc sulfide}$ as $279.2 \cdot m o l \times \frac{2}{3} = 186.1 \cdot m o l$; this quantity follows PRECISELY the stoichiometry of the given equation. Agreed?

i.e. $186.1 \cdot m o l \times 97.47 \cdot g \cdot m o {l}^{-} 1 = \text{over 18} \cdot k g$.