# Question #ae3df

##### 1 Answer
Apr 20, 2017

The reaction should be $4 A l + 3 {O}_{2} \to 2 A {l}_{2} {O}_{3}$

#### Explanation:

The molar ratio between $A l$ and ${O}_{2}$ is $4 \div 3$

$A l \left(m o l\right) \div 4.50 \left(m o l\right) = 4 \div 3$ or $\frac{A l}{4.50} = \frac{4}{3} \to A l = 6 \left(m o l\right)$

Check:
The molar ratio between $A l$ and $A {l}_{2} {O}_{3}$ is $2 \div 1$

So $6 m o l$ of $A l$ will give $6 \div 2 = 3 m o l$ of $A {l}_{2} {O}_{3}$