Question

Clay found in a lake at Bhopal contains 15% by mass moisture along with $CaC{O}_3$ and non volatile $Si{O}_2$. This on heating loses a part of its moisture, but $CaC{O}_3$ is completely converted into CaO. The partially dried sample now contains 7.3% moisture and 51.5% $Si{O}_2$. What is the mass percentage of $CaC{O}_3$ in the original sample of clay approximately?

Answer 1

Let % of $CaC{O}_3$ be x and we take 100 gm of clay then,
Mass of $Si{O}_2=85-x$
After heating $Si{O}_2$ is 51.5% and CaO is 41.2% as moisture is 7.3%
Since the mass of $Si{O}_2$ will remain same, then
$\frac{51.5}{100}\times weightoftheclayafterheating=85-x\therefore weightoftheclayafterheating=\frac{(85-x)100}{51.5}$.
So, mass of the $CaO$
$=\frac{(85-x)100}{51.5}\times \left(\frac{41.2}{100}\right)$ ---- (1)
Since $CaC{O}_3$ is completely converted to $CaO$, the number of moles of $CaO$ and $CaC{O}_3$ are the same.
So mass of $CaO$ will be $\frac{x}{100}\times 56$ ----(2)
From equations (1) and (2),
$⟹\frac{(85-x)100}{51.5}\times \left(\frac{41.2}{100}\right)=\frac{x}{100}\times 56$
$⟹x=50gm$