81 g of aluminium will react completely with how many grams of oxygen?

Steps to Solve

1. Identify the Balanced Chemical Equation

The reaction between aluminum (Al) and oxygen (O$_2$) can be represented by the chemical equation:

$$ 4 , \text{Al} + 3 , \text{O}_2 \rightarrow 2 , \text{Al}_2\text{O}_3 $$

This equation shows that 4 moles of aluminum react with 3 moles of oxygen.

2. Calculate the Molar Masses

Next, calculate the molar masses:

• Molar mass of aluminum (Al): approximately 27 g/mol
• Molar mass of oxygen (O$_2$): approximately 32 g/mol

3. Convert Grams of Aluminum to Moles

Now, convert the grams of aluminum into moles:

$$ \text{Moles of Al} = \frac{81 , \text{g}}{27 , \text{g/mol}} = 3 , \text{mol} $$

4. Use Stoichiometry to Find Moles of Oxygen

Using the stoichiometry from the balanced equation, find the moles of oxygen that will react:

From the equation, 4 moles of Al react with 3 moles of O$_2$. Therefore:

$$ \text{Moles of O}_2 = \frac{3}{4} \times \text{Moles of Al} = \frac{3}{4} \times 3 = 2.25 , \text{mol} $$

5. Convert Moles of Oxygen to Grams

Finally, convert the moles of oxygen back to grams:

$$ \text{Mass of O}_2 = \text{Moles of O}_2 \times \text{Molar mass of O}_2 $$

$$ \text{Mass of O}_2 = 2.25 , \text{mol} \times 32 , \text{g/mol} = 72 , \text{g} $$