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=
`(3)/(2)* 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=
`(2)/(3)* 2.85=1.9` moles of aluminium.

3 moles of oxygen give 2 moles of
`Al_2O_3`

2.85 moles of oxygen give=
`(2)/(3)* 2.85=1.9` moles of
`Al_2O_3`.

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