Question

27 g of Al will react completely with how much mass if O2 to produce Al2O3

Answer 1

the balanced reaction is

4Al + 3O₂ -----→ 2Al₂O₃
There for 4 mole of Al 3 moles of O₂

so for 1 mole Al 3/4 mole O₂

27 g Al means 1 mole so 3/4 mole O₂ is needed

ie 3/4 x 32 = 24 g

Answer 2

Given:

27 g of Al will react completely with oxygen gas to produce aluminium oxide.

To find:

We have to find the amount of oxygen used.

Solution:

The balanced equation of aluminium and oxygen reaction is as follows-

$4Al+3O_2\rightarrow 2Al_2O_3$.

Hence four moles of aluminium react with three moles of oxygen gas to form two moles of aluminium oxide.

Thus one mole of aluminium will react with 3/4 moles of oxygen gas to form aluminium oxide.

Now one mole of Al is equal to 27 grams of aluminium.

The molecular weight of oxygen gas is 32 grams.

Hence the amount of oxygen required is 3/4×32=24 grams.

Thus 24 grams of oxygen is required.