Question

Q13.How many moles of Al2O3 will be formed when a mixture of 5.4 g of Al and 3.2 g of O2 is heated ? [ Al = 27, 0 = 16]

Answer 1

Answer:

0.0666 moles of aluminum oxide will be formed.

Explanation:

$4Al+3O_2\rightarrow 2Al_2O_3$

Moles of aluminium = $\frac{5.4 g}{27 g/mol}=0.2 mol$

Moles of oxygen gas = $\frac{3.2 g}{16\times  g/mol}=0.1 mol$

According to reaction, 4 mole of aluminum reacts with 3 moles of oxygen gas.

Then 0.2 mol of aluminum will react with:

$\frac{3}{4}\times 0.2 mol=0.15 mol$ oxygen gas

Then 0.1 mol of oxygen will recat with:

$\frac{4}{3}\times 0.1 mol=0.13 mol$ of aluminum

As we can see that oxygen is in limiting amount. So the amount product formed will depend upon moles of oxygen gas.

According to reaction 3 moles of oxygen gas gives 2 moles of aluminium oxide.

Then 0.1 moles of oxygen will give:

$\frac{2}{3}\times 0.1 =0.0666 mol$ of aluminum oxide

0.0666 moles of aluminum oxide will be formed.

Answer 2

Answer:

1/15 or, 0.066

Explanation:

4Al+3O₂→2Al₂O₃

mass of Al=5.4ɡ

mass of O₂=3.2ɡ

now,moles of Al=5.4/27 (atomic wt.of Al=27)

=1/5 or,0.2

similarly,moles of O₂=3.2/32 (atomic wt.of O₂=32ɡ)

=1/10or,0.1

now,we have to find out the limitinɡ reaɡent.

So,for this we will divide the moles of the elements by stoichiometric coefficient

for O₂=0.1/3=0.033

ror Al=0.2/4=0.05

now,it is clear that O₂ is limitinɡ reaɡent ,as its value is less than Al

as per the reaction ,3moles of O₂ is required to make 2 mole of Al₂O₃

so,0.1 mole of O2 will make

2/3×0.1 mole of Al2O3

=1/15 =0.066