A 1.50 g sample of KHCO 3 having 80 % purity is strongly heated. Assuming the impurity to ste thermally stable, the loss in weight of the sample, on heating is:

Question

A 1.50 g sample of KHCO 3 having 80 % purity is strongly heated. Assuming the impurity to ste thermally stable, the loss in weight of the sample, on heating is: (A) 26.4 g (B) 2.64 g (C) 0.264 g (D) 0.0264 g

Tardigrade Admin · Accepted Answer

2 KHCO 3 xrightarrowΔ K 2 CO 3+ H 2 O + CO 2 uparrow Loss of wt. will be because of CO 2 escaped Total KHCO 3 chosen =1.50 g percentage purity =80 % ∴ Pure KHCO 3=1.50 × (80/100)=1.2 g M. mass of KHCO 3=39+1+12+3 × 16=100 g moles of pure KHCO 3=(1.2/100)=0.012 moles From balanced equation- 2 moles of KHCO 3 yield moles of CO 2=1 0.012 moles of KHCO 3 yield moles of CO 2=(1/2) × 0.012 =0.006 moles wt. of CO 2 formed =0.006 × 44=0.264 g

Q. A 1.50g sample of KHCO3​ having 80% purity is strongly heated. Assuming the impurity to ste thermally stable, the loss in weight of the sample, on heating is:

26.4 g

2.64 g

0.264 g

0.0264 g

Q. A $1.50 g$ sample of $KH C O_{3}$ having $80%$ purity is strongly heated. Assuming the impurity to ste thermally stable, the loss in weight of the sample, on heating is: