The theoretical yield of Aluminum oxide is 193.8 grams
calculation
Theoretical yield is calculate as below
the balanced equation for reaction is as below
4 Al +3 O2 → 2Al2O3
The mole ratio of Al: Al2O3 is 4 :2 therefore moles of Al2O3
= 3.40 x2/4 =1.7 moles
The mole ratio of O2 : Al2O3 is 3:2 therefore moles of Al2O3 is
= 2.85 x 2/3 =1.9 moles
N.b O2 is in excess therefore it's mole is used to determine moles of Al2O3 which is used to calculate theoretical yield.
theoretical yield is therefore = 1.9 moles x 102 g/mol= 193.8 grams
Answer: 193.8 grams
Explanation:
[tex]2Al+3O_2\rightarrow 2Al_2O_3[/tex]
As can be seen from the chemical equation, 2 moles of aluminium react with 3 moles of oxygen.
Thus 3.40 moles of aluminium react with=[tex]\frac{3}{2}\times 3.40=5.1[/tex] moles of oxygen.
But as the available amount of oxygen is less, oxygen is the limiting reagent as it limits the formation of product and aluminium is the excess reagent.
3 moles of oxygen react with 2 moles of aluminium
2.85 moles of oxygen react with=[tex]\frac{2}{3}\times 2.85=1.9[/tex] moles of aluminium.
3 moles of oxygen give 2 moles of [tex]Al_2O_3[/tex]
2.85 moles of oxygen give=[tex]\frac{2}{3}\times 2.85=1.9[/tex] moles of [tex]Al_2O_3[/tex].
Mas of [tex]Al_2O_3=moles\times {\text {molar mass}}=1.9\times 102g/mol=193.8g[/tex].
