The % loss in weight after heating a pure sample of potassium chlorate (M. wt. 122.5) will be- 2 KClO 3 stackrelΔ longrightarrow 2 KCl +3 O 2

Question

The % loss in weight after heating a pure sample of potassium chlorate (M. wt. 122.5) will be- 2 KClO 3 stackrelΔ longrightarrow 2 KCl +3 O 2 (A) 12.25 (B) 24.50 (C) 39.17 (D) 49.00

Tardigrade Admin · Accepted Answer

moles of KCl =2 Wt =150 Molar mass of KClO 3=122.5 g Mass of KCl =74.5 g % loss % loss =(122.5-74.5/122.5) × 100 =(48/122.5) × 100=39.18

Q. The $%$ loss in weight after heating a pure sample of potassium chlorate (M. wt. 122.5) will be-
$2 K Cl O_{3} ⟶ Δ 2 K Cl + 3 O_{2}$

$12.25$

$24.50$

$39.17$

$49.00$

moles of $K Cl = 2$

$W t = 150$

Molar mass of $K Cl O_{3} = 122.5 g$

Mass of $K Cl = 74.5 g$

$%$ loss

$%$ loss $= \frac{122.5 - 74.5}{122.5} \times 100$

$= \frac{48}{122.5} \times 100 = 39.18$

Q. The % loss in weight after heating a pure sample of potassium chlorate (M. wt. 122.5) will be- 2KClO3​⟶Δ​2KCl+3O2​

12.25

24.50

39.17

49.00