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

   315 kg

Explanation:

1. Word reaction:

Aluminum trioxide + hydrogen → aluminium metal + water

2. Chemical equation:

This is the unbalanced equation that represents chemical reaction:

       

   

3. Balanced equation

This is the balanced chemical equation that represents the reaction:

               

4. Calculate the number of moles of aluminium oxyde available:

   Molar mass = 101.960 g/mol    Mass = 700.0 kg

Formula:

Number of moles = mass in grams / molar mass

 

Substitute:

Number of moles = 700.0 kg × 1,000 g/ kg / 101.960 g/mol = 6,865.4mol

5. Mole ratio:

From the balanced chemical equation:

         

6. Number of moles of aluminium

       

7. Convert to mass

Atomic mass of aluminium: 26.981g/molMass = atomic mass × number of molesMass = 26.981g/mol ×  13,730.8mol = 370,470.7g

This is a theoretical yield.

8. Actual yield

Use the formula of percent yield.

Percent yield = (actual yield) / (theoretical yield) × 100

   

Actual yield = theoretical yield × percent yield / 100 Actual yield = 370,470.7 g × 85/100 = 314,900.1 g    314,900.1 g × 1kg/1,000g = 314.9kg = 315kg ← answer