Question

KClO_4 may be prepared by means of the following series of reactions:
Cl_2+2KOH→KCl+KClO+H_2 O
3KClO→2KCl+KClO_3
4KClO_3→3KClO_4+KCl
How much Cl2 is needed to prepare 200 g KClO_4 by the above sequence?

Answer 1

Cl2 + 2 KOH --------> KCl + KClO + H2O

3 KClO --------> 2 KCl + KClO3

4 KClO3 --------> 3 KClO4 + KCl

Before looking into the 1st equation,

we know,

22.4 litres of Cl2 indicates 1 mole of it

so, 4.48 litres of Cl2 indicates 0.2 moles

Also, number of moles of KOH = given mass/molar mass

= 11.2/56

= 0.2 moles

The 1st given equation is

Cl2 + 2 KOH --------> KCl + KClO + H2O

According to this equation, each mole of Cl2 requires 2 moles of KOH.

So, 0.2 moles of Cl2 requires 0.4 moles of KOH.

But there is only 0.2 moles of KOH present in the sample,

So, Only 0.1 moles of Cl2 will be able to participate in the reaction consuming all the 0.2 moles of KOH.

The other 0.1 moles of Cl2 left, will not have any KOH to react with! Hence it is the excess reagent which is left behind.

2) Mass of 0.1 moles Cl2 = 0.1 x 71 = 7.1 grams

So the reaction goes like this.

0.1 Cl2 + 0.2 KOH ------> 0.1 KCl + 0.1 KClO + 0.1 H2O

So, 0.1 moles of KClO is formed.

Now, 3 KClO --------> 2 KCl + KClO3

Accordingly, 3 moles of KClO produces 1 mole of KClO3

Calculating we get, 0.1 moles of KClO will produce 0.033 moles of KClO3

Coming to the third equation,

4 KClO3 --------> 3 KClO4 + KCl

4 moles of KClO3 gives 3 moles of KClO4

so, 0.033 moles of KClO3 will give 0.025 moles of KClO4