# Bauxite ore contains aluminum oxide, which is decomposed using electricity to produce aluminum metal. How many moles of aluminum metal can be produced if 100 g of oxygen is released?

Apr 1, 2016

You can produce $\text{8.33 mol of Al}$ from ${\text{100 g of O}}_{2}$.

#### Explanation:

There are three steps to answering this type of stoichiometry problem.

1. Write the balanced equation for the reaction.
2. Use the molar mass of ${\text{O}}_{2}$ to convert grams of ${\text{O}}_{2}$ to moles of ${\text{O}}_{2}$.
3. Use the molar ratio from the balanced equation to convert moles of ${\text{O}}_{2}$ to moles of $\text{Al}$.

Step 1. Write the balanced equation.

${\text{2Al"_2"O"_3 → "4Al" + "3O}}_{2}$

Step 2. Convert grams of ${\text{O}}_{2}$ to moles of ${\text{O}}_{2}$.

The molar mass of ${\text{O}}_{2}$ is 16.00 g/mol.

100 cancel("g O"_2) × ("1 mol O"_2)/(16.00 cancel("g O"_2)) = "6.250 mol O"_2

Step 3. Use the molar ratio to calculate the moles of $\text{Al}$.

From the balanced equation, the molar ratio is ${\text{4 mol Al":"3 mol O}}_{2}$.

6.250 cancel("mol O"_2) × ("4 mol Al")/(3 cancel("mol O"_2)) = "8.33 mol Al"

You can produce $\text{8.33 mol of Al}$.

Here's an interesting video on stoichiometry involving grams-to moles calculations.