Question

What is the theoretical yield of carbon monoxide that can be produced by the reaction of 60.0 g of aluminum oxide with 30.0 g of carbon according to the following chemical equation?Al2O3 + 3C → 2Al + 3CO

Answer 1

1) Write and balance the chemical equation

`Al_2O_3+3C\rightarrow2Al+3CO`

2) Calculate moles of aluminum oxide (Al2O3)

`\text{mol Al}_2O_3=60.0gAl_2O_3\cdot(1molAl_2O_3)/(101.96g)=0.5885molAl_2O_3`

3) Calculate moles of carbon (C)

`\text{mol C=30.0g C}\cdot\frac{1\text{ mol C}}{12.01\text{ g}}=2.498molC_{}`

4) Find the limiting reactant

How many moles of Al2O3 do we need to use all of the C.

`\text{molAl}_2O_3=2.498molC\cdot\frac{1molAl_2O_3}{3\text{ mol C}}=0.8327molAl_2O_3`

We need 0.8327mol Al2O3 and we have 0.5885 mol Al2O3. We do not have enough Al2O3. Al2O3 is the limiting reactant.

How many moles of C do we need to use all of the Al2O3.

`\text{mol C=0.5885 mol Al}_2O_3\cdot\frac{3\text{ mol C}}{1molAl_2O_3}=1.7655\text{ mol C}`

We need 1.7655 mol C and we have 2.498 mol C. We have enough C. This reactant is in excess.

5) Theoretical yield

The limiting reactant is Al2O3 (0.5885 mol)

`\text{mol CO=0.5885molAl}_2O_3\cdot\frac{3\text{ mol CO}}{1molAl_2O_3}=1.7655\text{ mol CO}`

6) grams of CO produced.

`g\text{ CO=}1.7655\text{molCO}\cdot\frac{28.01\text{ g CO}}{1\text{ mol CO}}=79.45\text{ g CO}`

The theoretical yield of carbon monoxide is 49.01 g.

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