Question

5. How many moles of Al Oz will be formed when a mixture of 5.4 g of Al and 3.2 g of O, is heated? 1 1 ( (2) 10 15 (1 ) → - 1 (3) (4) 20​

Answer 1

Answer:

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

27g/mol

5.4g

=0.2mol

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

16× g/mol

3.2g

=0.1mol

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

4

3

×0.2mol=0.15mol oxygen gas

Then 0.1 mol of oxygen will recat with:

\frac{4}{3}\times 0.1 mol=0.13 mol

3

4

×0.1mol=0.13mol 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

3

2

×0.1=0.0666mol of aluminum oxide

0.0666 moles of aluminum oxide will be formed.

Answer 2

Answer:

0.067

Explanation:

The required reaction is:-

Al+O2 - Al2O3

First we neet to find the limitiong reagent for that first balance the equation

4Al+ 3O2- 2Al2O3

Now find moles of Al and O2

Moles of Al= 5.4/27=0.2 moles

Moles of O2= 3.2/32=0.1 moles

To find mole ratio divide moles with stoichiometric coefficient

Mole ratio of al=0.2/4 =0.05

Moles ratio of O2=0.1/3=0.333

So O2 is limiting reagent.

So we calculate moles of al2o3 from O2.

1 moles of O2 gives 2/3 moles of Al2O3

So 0.1 moles woukd give 2/3×1/10=2/30=0.067 mmoles