If $50 \cdot g$ $50*g$ of aluminum is oxidized, what mass of alumina will be recovered?

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If $50 \cdot g$ $50*g$ of aluminum is oxidized, what mass of alumina will be recovered?

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Answer 1 · 2017-07-24T21:19:24

Approx. $95 \cdot g$ $95*g$ of alumina will be recovered.

Explanation:

We follow the stoichiometric equation........

$2 A l (s) + \frac{3}{2} O_{2} (g) \to A l_{2} O_{3} (s)$ $2Al(s) + 3/2O_2(g) rarr Al_2O_3(s)$

And this 2 equiv of metal required 3/2 equivalents of dioxygen gas.

$Moles of aluminum$ $"Moles of aluminum"$ $=$ $=$ $\frac{50 \cdot g}{27.0 \cdot g \cdot m o l^{- 1}} = 1.85 \cdot m o l$ $(50*g)/(27.0*g*mol^-1)=1.85*mol$ .

And since aluminium is CLEARLY the so-called limiting reagent in this reaction (since we know precisely how much mass of $A l$ $Al$ reacted), at most we can get half an equiv of alumina....

And thus the mass of alumina is given by the product....

$1.85 \cdot m o l \times \frac{1}{2} \times 101.96 \cdot g \cdot m o l^{- 1} = ? ? \cdot g$ $1.85*molxx1/2xx101.96*g*mol^-1=??*g$ .

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If $50 \cdot g$ $50*g$ of aluminum is oxidized, what mass of alumina will be recovered?

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If $50 \cdot g$ $50*g$ of aluminum is oxidized, what mass of alumina will be recovered?