Respuesta :
1) Write and balance the chemical equation
[tex]Al_2O_3+3C\rightarrow2Al+3CO[/tex]2) Calculate moles of aluminum oxide (Al2O3)
[tex]\text{mol Al}_2O_3=60.0gAl_2O_3\cdot\frac{1molAl_2O_3}{101.96g}=0.5885molAl_2O_3[/tex]3) Calculate moles of carbon (C)
[tex]\text{mol C=30.0g C}\cdot\frac{1\text{ mol C}}{12.01\text{ g}}=2.498molC_{}[/tex]4) Find the limiting reactant
How many moles of Al2O3 do we need to use all of the C.
[tex]\text{molAl}_2O_3=2.498molC\cdot\frac{1molAl_2O_3}{3\text{ mol C}}=0.8327molAl_2O_3[/tex]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.
[tex]\text{mol C=0.5885 mol Al}_2O_3\cdot\frac{3\text{ mol C}}{1molAl_2O_3}=1.7655\text{ mol C}[/tex]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)
[tex]\text{mol CO=0.5885molAl}_2O_3\cdot\frac{3\text{ mol CO}}{1molAl_2O_3}=1.7655\text{ mol CO}[/tex]6) grams of CO produced.
[tex]g\text{ CO=}1.7655\text{molCO}\cdot\frac{28.01\text{ g CO}}{1\text{ mol CO}}=79.45\text{ g CO}[/tex]The theoretical yield of carbon monoxide is 49.01 g.
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