Question

One mole of potassium chlorate is thermally decomposed and excess of aluminum is
burnt in gascous mixture. How many moles of aluminum oxide are formed?
Respective equations are given as
2KCIO3 ---->KC1+302, 4A1+30, ------> 2Al203
C) 1.5
B) 2.0
D) 3.0
A) 1.0​

Answer 1

Answer:

Explanation:

The thermal decomposition of KClO3 and the subsequent oxidation of Aluminum to form Al2O3.

Two reactions are: 

2 KClO3 → 2 KCl + 3 O2 and 4 Al + 3 O2 → 2 Al2O3 

The molar ratio for the first reaction is 2:3 for KClO3: O2. Thus, if 1 mole KClO3 is decomposed, 1.5 moles of O2 is produced ===> 1 mol KClO3 x ( 3 mol O2 / 2 mol KClO3) = 1.5 mol O2 

The molar ratio for the second reaction is: 3:2 for O2 : Al2O3. ===> 3 moles of oxygen produces 2 moles of aluminum oxide.

Since we have 1.5 moles of oxygen (half of the three needed) we can expect to produce half of the 2 moles or 1 mole aluminum oxide. Hence, 

1.5 mol O2 x ( 2 mol Al2O3 / 3 mol O2) = 1 mol of  Al2O3...........

Option A is correct!