Question

According to the following reaction, how many grams of aluminum oxide will be formed upon the complete reaction of 21.4 grams of aluminum with excess iron(III) oxide?

iron(III) oxide (s) + aluminum (s) aluminum oxide (s) + iron (s)------ grams aluminum oxide

Answer 1

To find the mass of aluminum oxide formed, we need to calculate the number of moles of aluminum, use the stoichiometry of the reaction to find the moles of aluminum oxide produced, and then convert those moles to mass using the molar mass of aluminum oxide.

Step-by-step explanation:

The balanced chemical equation for the reaction between aluminum (Al) and iron(III) oxide (Fe₂O₃) is 2 Al(s) + Fe₂O₃(s) → 2 Fe(s) + Al₂O₃(s). To determine how many grams of aluminum oxide (Al₂O₃) will be formed from 21.4 grams of aluminum reacting with excess iron(III) oxide, we must follow these steps:

1. Calculate the molar mass of Al, which is approximately 26.98 g/mol.
2. Determine the number of moles of Al in 21.4 grams by dividing mass by molar mass (21.4 g Al / 26.98 g/mol Al).
3. Use the stoichiometry of the balanced equation, which shows that 2 moles of Al produce 1 mole of Al₂O₃.
4. Calculate the molar mass of Al₂O₃, which is approximately 101.96 g/mol.
5. Multiply the number of moles of Al₂O₃ by its molar mass to get the mass of Al₂O₃ produced.

Following these steps will allow us to compute the exact mass of Al₂O₃ formed.