Given the balanced equation:
2 Al + Fe2O3 → Al2O3 + 2 Fe
how many grams of Al2O3 will be produced if 68.1 g Al reacts with 112 g Fe2O3?
theoretical yield problem. Rather than finding Limiting reagent by getting to moles, lets do one full step becuase we are asked for grams of Al2O3.
68.1 g Al/ molar mass = mol Al times mole ratio times molar mass Al2O3. Same procedure for Fe2O3. Produced less is theoretical yield(you know that was your LR) and that is amount Al2O3 in grams produced.
To determine the grams of Al2O3 produced, we need to use stoichiometry, which involves converting the given amounts of reactants (Al and Fe2O3) to the desired product (Al2O3) using the coefficients from the balanced equation.
Step 1: Calculate the number of moles of Al using its molar mass.
Molar mass of Al = 26.98 g/mol
Number of moles of Al = 68.1 g Al / 26.98 g/mol = 2.52 mol Al
Step 2: Calculate the number of moles of Fe2O3 using its molar mass.
Molar mass of Fe2O3 = 55.85 g/mol (Fe) + 16.00 g/mol (O) × 3 = 159.69 g/mol
Number of moles of Fe2O3 = 112 g Fe2O3 / 159.69 g/mol = 0.70 mol Fe2O3
Step 3: Use the stoichiometric ratio from the balanced equation to determine the limiting reactant. The stoichiometric ratio between Al and Al2O3 is 2:1. Therefore, the number of moles of Al2O3 produced will be half the number of moles of Al, assuming all the Al reacts completely.
Number of moles of Al2O3 produced = 0.5 × 2.52 mol Al = 1.26 mol Al2O3
Step 4: Calculate the grams of Al2O3 produced using its molar mass.
Molar mass of Al2O3 = 26.98 g/mol (Al) × 2 + 16.00 g/mol (O) × 3 = 101.96 g/mol
Grams of Al2O3 produced = 1.26 mol Al2O3 × 101.96 g/mol = 128.67 g Al2O3
Therefore, if 68.1 g Al reacts with 112 g Fe2O3, approximately 128.67 g of Al2O3 will be produced.