Question

How much potassium chloride, KCl, is produced during the decomposition of 100.0 grams of potassium chlorate, KClO3?

Answer 1

Answer : The correct answer is : 61.13 g

Potassium chlorate decompose to give potassium chloride and oxygen . The balanced reaction is as follows :

2 KClO₃ → 2 KCl + 3 O₂

Given : Mass of KClO₃ = 100.0 g Mass of KCl = ?

Following are the steps to calculate mass of KCl :

Step 1: Convert mass of KClO₃ to its mole .

`Mole = (Mass)/(Molar mass )`

Molar mass of KClO₃ = 122.55
`(g)/(mol)`

`Mole = (100 g)/( 122.55 (g)/(mol)) = (100 g)/(122.55 g) * 1 mol`

Mole = 0.82 mol

Step 2: Find mole ratio between moles of KClO₃ and KCl .

In balanced reaction the coefficients of KClO₃ is 2 and that of KCl is also 2. Hence the mole ratio of KClO₃ : KCl = 2 : 2 = 1 : 1 .

Hence mole of KCl =>

`0.82 mole of KClO3 * (1 mol KCl)/(KClO3)` = 0.82 mol of KCl .

Step 3: Conversion of mole into grams .

`Mole = (Mass)/(Molar mass )`

Molar mass of KCl = 74.55
`(g)/(mol)`

Plugging value in above formula :

`0.82 mol KCl = ( mass of KCl )/(74.55 (g)/(mol))`

Multiplying both side by
`{74.55 (g)/(mol)}` :

`0.82 mol KCl * {74.55 (g)/(mol) = ( mass of KCl )/(74.55 (g)/(mol)) * {74.55 (g)/(mol)`

Mass of KCl = 61.13 g

Hence mass of KCl produced = 61.13 g