# How much dioxygen is required to oxidize a 4*mol quantity of aluminum metal?

Dec 30, 2016

If the reaction involved is this one

$4 A l \left(s\right) + 3 {O}_{2} \left(g\right) \rightarrow 2 A {l}_{2} {O}_{3} \left(s\right)$

the answer is 11.1 moles of ${O}_{2}$

#### Explanation:

I'm going to assume that the reaction you are referring to is

$4 A l \left(s\right) + 3 {O}_{2} \left(g\right) \rightarrow 2 A {l}_{2} {O}_{3} \left(s\right)$

In which case, the ratio of $\left(\text{moles of "O_2 " required")/("moles of "Al " used}\right) = \frac{3}{4}$

Write an equation that has this ratio equal to a similar one in which you use the numbers given in this problem, with $x$ representing the unknown amount of ${O}_{2}$:

$\left(\text{moles of "O_2 " required")/("moles of "Al " used}\right) = \frac{3}{4} = \frac{x}{14.8}$

Solve for $x$

$x = \left(\frac{3}{4}\right) \left(14.8\right) = 11.1 \text{ moles of } {O}_{2}$

Dec 30, 2016

We need $11.1$ moles of ${O}_{2} \left(g\right)$.

#### Explanation:

A stoichiometric equation is required:

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

Given the equation, for each equiv metal $\frac{3}{4}$ equiv dioxygen gas are required, i.e. $14.8 \cdot m o l \times \frac{3}{4} = 11.1 \cdot \text{mol}$ ${O}_{2}$ gas.

Given standard condtions, how many litres of gas does this represent?