In a reaction, 34.6 g of Cr2O3 reacts with 12.3 g of Al to produce cr and Al2O3 If 23.7 g of Cr is produced, what mass of Al2O3 is produced?
To solve this problem, we need to set up a balanced equation for the reaction and use stoichiometry to calculate the mass of Al2O3 produced.
The balanced equation for the reaction between Cr2O3 and Al is:
2Cr2O3 + 6Al → 4Cr + 3Al2O3
From the balanced equation, we see that 2 moles of Cr2O3 react with 6 moles of Al to produce 3 moles of Al2O3.
First, let's calculate the number of moles of Cr2O3 and Al using their respective masses:
Mass of Cr2O3 = 34.6 g
Molar mass of Cr2O3 = 2 * (Cr atomic mass) + 3 * (O atomic mass) = 2 * (52 g/mol) + 3 * (16 g/mol) = 152 g/mol
Number of moles of Cr2O3 = mass / molar mass = 34.6 g / 152 g/mol = 0.2276 mol
Mass of Al = 12.3 g
Molar mass of Al = 26.98 g/mol
Number of moles of Al = mass / molar mass = 12.3 g / 26.98 g/mol = 0.456 mol
Now, let's determine the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.
From the balanced equation, we see that 2 moles of Cr2O3 react with 6 moles of Al to produce 3 moles of Al2O3.
The stoichiometric ratio between Cr2O3 and Al is 2:6, or simplified, 1:3. Therefore, for every mole of Cr2O3, 3 moles of Al are required.
Since the number of moles of Cr2O3 (0.2276 mol) is less than three times the number of moles of Al (3 * 0.456 mol = 1.368 mol), Al is the limiting reactant.
Now we can calculate the mass of Al2O3 produced using the stoichiometry of the balanced equation.
Molar mass of Al2O3 = 2 * (Al atomic mass) + 3 * (O atomic mass) = 2 * (26.98 g/mol) + 3 * (16 g/mol) = 101.96 g/mol
Number of moles of Al2O3 produced = Number of moles of limiting reactant (Al) * (3 moles of Al2O3 / 6 moles of Al) = 0.456 mol * (3/6) = 0.228 mol
Mass of Al2O3 produced = Number of moles of Al2O3 produced * molar mass of Al2O3 = 0.228 mol * 101.96 g/mol = 23.3 g
Therefore, the mass of Al2O3 produced is 23.3 g.