Question

A 1.5g sample of potassium bicarbonate having 80% purity is strongly heated.Assuming the impurity to be thermally stable,the loss in weight of the sample,on heating ,is_________

Answer 1

Answer: loss in weight of sample = 0.372 g

2 kHCO3 -> K2 CO3 + H2O + CO2

200                138           18        44

on strong heating h2o and co2 will escape but K2CO3 WILL NOT decompose as it is stable

Explanation:

80 % Of 1.5 g means 80 x 1.5 /100 = 1.2 g that means in 1.5 g 1.2 g only is potassium bicarbonate remaining is impurity

from above equation

200 g on heating will give 18 + 44 = 62g

1.2 g will give 0.372 g

so loss in weight = 0.372 g

Answer 2

Answer:

Explanation:2 kHCO3 -> K2 CO3 + H2O + CO2

200                138           18        44

on strong heating h2o and co2 will escape but K2CO3 WILL NOT decompose as it is stable

Explanation:

80 % Of 1.5 g means 80 x 1.5 /100 = 1.2 g that means in 1.5 g 1.2 g only is potassium bicarbonate remaining is impurity

from above equation

200 g on heating will give 18 + 44 = 62g

1.2 g will give 0.372 g

so loss in weight = 0.372 g

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