Question

$4.90 \,g$ of impure potassium chlorate on heating shows a weight loss of $0.384\, g$. What per cent of the impure potassium chlorate has decomposed?

• A. 20
• B. 30
• C. 40
• D. 80

Answer 1

Answer: (A)

Given,

Impure potassium chlorate $=4.90\, g$

On heating, weight loss $=0.384\, g$

Equation

$2 KClO _{3} \ce{->[{\Delta}]} 2 KCl +3 O _{2}$

$2 $ mol of $KClO _{3}$ will produce $2$ mol $KCl$ and $3$ mol $O _{2}$

Molar mass of $KClO _{3}=122.55 \,g / mol$

Molar mass of $O _{2}=32 \,g / mol$

$0.384 \,g$ loss$=0.384 \,g O _{2}$ produced

So, produced $O _{2}=\frac{0.384}{32}=0.012$ mol

That means, decomposed

$KClO _{3}=0.012 \times 2 / 3=0.008$

Mass of decomposed

$KClO _{3}=122.55 \times 0.008=0.980\, g$

$\%$ of decomposed $KClO _{3}=\frac{0.980}{4.9} \times 100=20 \%$