Question

81 g of Al reacts with 128 g of \( O_2 \). Calculate the amount of \( Al_2O_3 \) produced.

Answer 1

The balanced chemical equation for the reaction is: \[ 4 \, \text{Al} + 3 \, \text{O}_2 \rightarrow 2 \, \text{Al}_2\text{O}_3. \] Step 1: Find the molar masses of aluminum (Al) and oxygen (O₂): - Molar mass of Al = 27 g/mol, - Molar mass of O₂ = 32 g/mol. Step 2: Convert the given masses to moles: - Moles of Al = \( \frac{81}{27} = 3 \, \text{mol} \), - Moles of O₂ = \( \frac{128}{32} = 4 \, \text{mol} \). Step 3: Use the stoichiometric ratio from the balanced equation: From the equation, 4 moles of Al react with 3 moles of O₂ to produce 2 moles of Al₂O₃. So, the moles of Al₂O₃ produced can be found by the ratio: \[ \text{Moles of } Al_2O_3 = \frac{3}{4} \times 3 = 2.25 \, \text{mol}. \] Step 4: Convert the moles of Al₂O₃ to grams: - Molar mass of Al₂O₃ = \( 2 \times 27 + 3 \times 16 = 102 \, \text{g/mol} \), - Mass of Al₂O₃ = \( 2.25 \times 102 = 229.5 \, \text{g} \). Thus, the amount of \( Al_2O_3 \) produced is \( 229.5 \, \text{g} \).