Question

heatWhat is the theoretical yield of aluminum oxide if 3.40 mol of aluminum metal is exposed to 2.85 mol of oxygen?

Answer 1

Answer: 193.8 grams

Explanation:

$2Al+3O_2\rightarrow 2Al_2O_3$

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=$\frac{3}{2}\times 3.40=5.1$ 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=$\frac{2}{3}\times 2.85=1.9$ moles of aluminium.

3 moles of oxygen give 2 moles of $Al_2O_3$

2.85 moles of oxygen give=$\frac{2}{3}\times 2.85=1.9$ moles of $Al_2O_3$.

Mas of $Al_2O_3=moles\times {\text {molar mass}}=1.9\times 102g/mol=193.8g$.

Answer 2

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