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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

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Edufirst

Answer:

  •    315 kg

Explanation:

1. Word reaction:

  • Aluminum trioxide + hydrogen → aluminium metal + water

2. Chemical equation:

This is the unbalanced equation that represents chemical reaction:

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

   

3. Balanced equation

This is the balanced chemical equation that represents the reaction:

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

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:

          [tex]\dfrac{2molAl}{1molAl_2O_3}[/tex]

6. Number of moles of aluminium

       [tex]6,865.4molAl_2O_2\times \dfrac{2molAl}{1molAl_2O_3}=13,730.8molAl[/tex]

7. Convert to mass

  • Atomic mass of aluminium: 26.981g/mol

  • Mass = atomic mass × number of moles
  • Mass = 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