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

Answer 1

Answer: (C)

$2 KHCO _{3} \xrightarrow{\Delta} K _{2} CO _{3}+ H _{2} O + CO _{2} \uparrow$

Loss of wt. will be because of $CO _{2}$ escaped

Total $KHCO _{3}$ chosen $=1.50 \,g$

percentage purity $=80 \%$

$\therefore $ Pure $KHCO _{3}=1.50 \times \frac{80}{100}=1.2\, g$

M. mass of $KHCO _{3}=39+1+12+3 \times 16=100 \,g$

moles of pure $KHCO _{3}=\frac{1.2}{100}=0.012$ moles

From balanced equation-

2 moles of $KHCO _{3}$ yield moles of $CO _{2}=1$

0.012 moles of $KHCO _{3}$ yield moles of $CO _{2}=\frac{1}{2} \times 0.012$

$=0.006$ moles

wt. of $CO _{2}$ formed $=0.006 \times 44=0.264 \,g$