# How many moles of Al_2O_3 can be prepared from 11*mol of aluminum?

May 24, 2017

$5.5 \cdot m o l \times 101.96 \cdot g \cdot m o {l}^{-} 1 = 560.78 \cdot g$...........

#### Explanation:

If I have 18 eggs, then how many dozen eggs do I have? I think you would quickly be able to respond that you have $\frac{3}{2}$ $\text{dozen}$, because $\text{1 dozen}$ $=$ $12$.

If I have a mole of stuff, aluminum atoms, oxygen atoms, I have $\text{Avogadro's number}$ of individual atoms, where $\text{Avogadro's number} \equiv 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

Here we have $11 \cdot m o l$ of aluminum metal; given the stoichiometry we can make $5.5 \cdot m o l$ $A {l}_{2} {O}_{3}$, which has a molar mass of $101.96 \cdot g \cdot m o {l}^{-} 1$.

And thus we take the product, $\text{molar quantity"xx"molar mass}$ to get an actual mass.