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

Jul 24, 2017

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

#### Explanation:

$2 A l \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) \rightarrow A {l}_{2} {O}_{3} \left(s\right)$

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

$\text{Moles of aluminum}$ $=$ $\frac{50 \cdot g}{27.0 \cdot g \cdot m o {l}^{-} 1} = 1.85 \cdot m o l$.

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

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

1.85*molxx1/2xx101.96*g*mol^-1=??*g.

Look for videos on molar equivalence and stoichiometry on the web. Perhaps this one: