Question

How much aluminum oxide and how much carbon are needed to prepare 681 g of aluminum by the balanced chemical reaction (below) if the reaction proceeds to 79.5% yield?

(Provide the balanced chemical reaction and compute the required amounts)

Answer 1

To produce 681 g of aluminum at a 79.5% yield, 539.48 g of aluminum oxide and 47.67 g of carbon are necessary, calculated using the balanced chemical equation for aluminum production.

Step-by-step explanation:

The student is asking how much aluminum oxide (Al2O3) and carbon (C) are needed to produce 681 g of aluminum (Al) based on a certain chemical reaction with a 79.5% yield. The balanced chemical equation for the production of aluminum from aluminum oxide is:

4 Al2O3(s) + 3 C(s) → 6 Al(l) + 3 CO2(g)

To calculate the required amounts, follow these steps:

1. Determine the moles of aluminum produced: 681 g Al × (1 mol Al / 26.98 g Al) = 25.24 mol Al.
2. Calculate the moles of aluminum that would be needed based on 100% yield: 25.24 mol Al × (1/6) = 4.207 mol Al2O3 (from the reaction stoichiometry).
3. Account for the actual yield of 79.5%: 4.207 mol Al2O3 × (1/0.795) = 5.29 mol Al2O3.
4. Determine the mass of Al2O3 needed: 5.29 mol Al2O3 × (101.96 g Al2O3 / mol) = 539.48 g Al2O3.
5. Determine moles of carbon needed: 5.29 mol Al2O3 × (3/4) = 3.97 mol C.
6. Determine the mass of carbon needed: 3.97 mol C × (12.01 g C / mol) = 47.67 g C.

In conclusion, to prepare 681 g of aluminum at a 79.5% yield, 539.48 g of aluminum oxide and 47.67 g of carbon are required.