Question

When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is ----- (nearest integer).

Hint:

Use stoichiometry to calculate the mass of the product based on the limiting reagent in the reaction.

Answer 1

Correct Answer: 153

The reaction between aluminium and oxygen is: \[ 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3. \] - Moles of aluminium = \( \frac{81.0}{27.0} = 3 \text{ moles} \). - Moles of oxygen = \( \frac{128.0}{32.0} = 4 \text{ moles} \). From the stoichiometry of the reaction, 4 moles of aluminium reacts with 3 moles of oxygen to produce 2 moles of aluminium oxide. Thus, 3 moles of aluminium will produce: \[ \frac{2}{4} \times 3 = 1.5 \text{ moles of } \text{Al}_2\text{O}_3. \] The molar mass of aluminium oxide \( \text{Al}_2\text{O}_3 \) is: \[ \text{Molar mass of Al}_2\text{O}_3 = 2(27.0) + 3(16.0) = 102.0 \text{ g/mol}. \] Thus, the mass of aluminium oxide produced is: \[ 1.5 \times 102.0 = 153 \text{ g}. \] The mass of aluminium oxide produced is 153 g.