In a reaction, 34.6 g of chromium(III) oxide reacts with 12.3 g of aluminum to produce chromium and aluminum oxide. If 23.7 g of chromium is produced, what mass of aluminum oxide is produced?
To calculate the mass of aluminum oxide produced, we need to first determine the chemical equation for the reaction and the stoichiometric ratios.
The balanced chemical equation for the reaction is:
2 Cr2O3 + 6 Al → 4 Cr + 3 Al2O3
From the balanced equation, we can see that 2 moles of chromium(III) oxide react with 6 moles of aluminum to produce 4 moles of chromium and 3 moles of aluminum oxide.
1. Convert the given mass of chromium(III) oxide to moles using its molar mass. The molar mass of Cr2O3 is:
(2 × atomic mass of Cr) + (3 × atomic mass of O)
= (2 × 52.0 g/mol) + (3 × 16.0 g/mol)
= 52.0 g/mol + 48.0 g/mol
= 100.0 g/mol
Number of moles of Cr2O3 = mass of Cr2O3 / molar mass of Cr2O3
= 34.6 g / 100.0 g/mol
= 0.346 mol
2. Determine the number of moles of aluminum using its molar mass. The molar mass of Al is 26.98 g/mol.
Number of moles of Al = mass of Al / molar mass of Al
= 12.3 g / 26.98 g/mol
= 0.456 mol
3. Calculate the limiting reactant. The limiting reactant is the reactant that is completely consumed, limiting the amount of product that can be formed. It can be determined by comparing the stoichiometric ratios of the reactants in the balanced equation.
Since 2 moles of Cr2O3 react with 6 moles of Al, the stoichiometric ratio of Cr2O3 to Al is 2:6, which simplifies to 1:3. Therefore, 1 mole of Cr2O3 requires 3 moles of Al.
Given that we have 0.346 mol of Cr2O3 and 0.456 mol of Al, the Al is the limiting reactant because it is in deficit compared to the 1:3 stoichiometric ratio.
4. Calculate the number of moles of chromium produced using the stoichiometry between Cr2O3 and Cr:
Number of moles of Cr = number of moles of Cr2O3 × (4 mol Cr / 2 mol Cr2O3)
= 0.346 mol × (4 mol Cr / 2 mol Cr2O3)
= 0.692 mol Cr
5. Finally, determine the mass of aluminum oxide produced using the stoichiometry between Al2O3 and Cr:
Mass of Al2O3 = number of moles of Cr × (3 mol Al2O3 / 4 mol Cr)
= 0.692 mol × (3 mol Al2O3 / 4 mol Cr)
= 0.519 mol Al2O3
To calculate the mass, multiply the number of moles of Al2O3 by its molar mass. The molar mass of Al2O3 is (2 × atomic mass of Al) + (3 × atomic mass of O) = (2 × 26.98 g/mol) + (3 × 16.0 g/mol) = 101.96 g/mol.
Mass of Al2O3 = number of moles of Al2O3 × molar mass of Al2O3
= 0.519 mol × 101.96 g/mol
= 52.99 g
Therefore, approximately 52.99 g of aluminum oxide is produced.