Question

How many moles of al2o3 will be formed when a mixture of 5.4 gram of al and 3.2 gram of o2 is heated?

Answer 1

Answer:

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

We know that the moles of aluminium = 5.4/27 = 0.1 mole..

Moles of oxygen gas = 3.2/(16*2) = 0.1 mole..

So, from the reaction we get 4 moles of Aluminium reacts with 3 moles of oxygen.  

From the equation we know that 0.2 mol of aluminum will react with:

We get that 3/4*0.2 = 0.15 mol of the  oxygen gas.

Consequently 0.1 mol of oxygen will react with:

So, we get that 4/3*0.1=0.13 mol of aluminum..

We can conclude that oxygen is the limiting reagent.

Since we know that 2 moles oxygen gives 3 moles Al2O3 then 0.1 moles of oxygen will give 2/3*0.1= 0.066 of aluminum oxide.

Answer 2

Answer:

1/15 or 0.066 mol

Explanation: 2Al+3/2O2 = Al2O3

Given: 5.4gAl and 3.2g O2

Moles of Al= 5.4/54=0.1

Moles of O2 = 3.2/48= 1/15=0.066

Moles of O2 are limiting reagent

Product formation depends upon L.R.

Moles of Al2O3 will form =1/15=0.066