Question

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

Answer 1

Answer:

• 315 kg

Explanation:

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:

     $Al_2O_3+3H_2\rightarrow 2Al+3H_2O$

2. Mole ratios

      $1molAl_2O_3:3molH_2:2molAl:3molH_2O$

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 $1molAl_2O_3:2molAl$; 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