How many grams of $O_{2}$ $O_2$ are required to produce 358.5 grams of $Z n O$ $ZnO$ in $2 Z n + O_{2} \to 2 Z n O$ $2Zn + O_2 -> 2ZnO$ ?

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How many grams of $O_{2}$ $O_2$ are required to produce 358.5 grams of $Z n O$ $ZnO$ in $2 Z n + O_{2} \to 2 Z n O$ $2Zn + O_2 -> 2ZnO$ ?

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Answer 1 · 2017-03-09T22:05:02

We have the stoichiometric equation.........

Explanation:

$Z n (s) + \frac{1}{2} O_{2} (g) \to Z n O (s)$ $Zn(s) + 1/2O_2(g) rarr ZnO(s)$

And we have $\frac{358.5 \cdot g}{81.41 \cdot g \cdot m o l^{- 1}} = 4.40 \cdot m o l$ $(358.5*g)/(81.41*g*mol^-1)=4.40*mol$ of zinc oxide.

Given the stoichiometry, $2.20 \cdot m o l$ $2.20*mol$ of dioxygen gas were necessary, i.e. $2.20 \cdot m o l \times 32.00 \cdot g \cdot m o l^{- 1} = 70.0 \cdot g$ $2.20*molxx32.00*g*mol^-1=70.0*g$

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How many grams of $O_{2}$ $O_2$ are required to produce 358.5 grams of $Z n O$ $ZnO$ in $2 Z n + O_{2} \to 2 Z n O$ $2Zn + O_2 -> 2ZnO$ ?

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Explanation:

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How many grams of O2O_2 are required to produce 358.5 grams of ZnOZnO in 2Zn+O2→2ZnO2Zn + O_2 -> 2ZnO?

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Explanation:

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How many grams of $O_{2}$ $O_2$ are required to produce 358.5 grams of $Z n O$ $ZnO$ in $2 Z n + O_{2} \to 2 Z n O$ $2Zn + O_2 -> 2ZnO$ ?