Question

How many moles of al2o3 are formed from when a mixture of 5.4g of al and 3.2g of o3 heated?

Answer 1

Answer:

0.1 moles of Al2O3

Explanation:

In order to determine the number of moles of Al2O3 produced when 5.4g of Al and 3.2g of O3 reacts, we need to write a balanced equation for the reaction:

2Al(s)+ O3(g) )⟶ Al2O3(s)

The number of moles of Al = (Mass of Al)/(RFM of Al)

= 5.4g/(27gmol-1)

= 0.2 moles of Al

From the mole ratio: Al: Al2O3

           2:  1

Moles of Al2O3= 1/2×0.2moles

= 0.1 moles of Al2O3

Answer 2

Answer:

From the equation we get that 4Al + 3O2 -> 2Al2O3.

We know that the moles of aluminium will be = 5.4/27 = 0.2mol.

And, moles of oxygen gas will be  = 3.2/32 = 0.1mol.

From the above equation we get to know that 3 moles of oxygen atom reacts with 4 moles of aluminium atom to give 2 moles aluminium oxide.

Then the limiting reagent will be oxygen atom itself.

As we got that 0.1 moles of oxygen will evolve then to get the aluminium oxide we know that 2 moles oxygen reacts to form 3 moles of aluminum oxide.

Then 2/3 moles will give 2/3*0.1 moles of aluminum oxide.

Which is 0.0666 moles of aluminum oxide will be formed.