A process to produce aluminum from aluminum oxide has an 85.0% yield. How much aluminum will be produced from a reaction 700.0 kg of aluminum oxide to produce Al? (assume that the reaction is Al2O3 + H2=Al+H2O

First you need to calculate the theoretical yield using the balanced chemical equation and the amount of aluminium available to react. Next, you use the percent yield to calculate the true yield.

1. Balanced chemical equation:

[tex]Al_2O_3+3H_2\rightarrow 2Al+3H_2O[/tex]

2. Mole ratios

[tex]1molAl_2O_3:3molH_2:2molAl:3molH_2O[/tex]

3. Number of moles of aluminium oxide

Molar mass = 101.960 g/mol
Mass = 700.0 kg = 7.0000 × 10⁵ g

Number of moles = mass in grams / molar mass
Number of moles = 7.0000 × 10⁵ g / 101.960 g/mol = 6,865.4mol

4. Number of moles of aluminium

The mole ratio is [tex]1molAl_2O_3:2molAl[/tex]; thus 6,865.4 mol of aluminium oxide produce 2 × 6,865.4 mol of aluminum = 13,730.8 mol of aluminium.

5. Calculate the mass of 13,730.8 mol of aluminium

Atomic mass of aluminium: 26.981g/mol

Mass = atomic mass × number of moles
Mass = 26.981g/mol × 13,730.8mol = 370,470.7g

6. Calculate the actual yield

Actual yield = theoretical yiel × percent yield

Actual yield = 85.0% × 370,470.7 g = 314,900.1 g

7. Convert to kg and three significant figures

314,900.1 g × 1kg/1,000g = 314.9kg = 315kg