# How many moles of "Al" are needed to form 3.7 moles of "Al"_2"O"_3?

## $4 {\text{Al" + 3"O"_2 -> 2"Al"_2"O}}_{3}$

Feb 20, 2018

$\text{7.4 mol Al}$

#### Explanation:

The balanced chemical equation for it is

$4 A l + 3 {O}_{2} \to 2 A {l}_{2} {O}_{3}$

You have $3.7$ moles of $A {l}_{2} {O}_{3}$. When you balanced the equation, you got $2$ moles of "Al_2O_3 and $4$ moles of $A l$, so

"4 mol $A l$ gives $2$ mol $A {l}_{2} {O}_{3}$

After you cancel out the units of measurement to check the order of the conversion, multiply $3.7$ with $4$ and then divide the answer by $2$. You get $7.4$ moles of $A l$.