TEST SERIES (CHEMISTRY FST)
263893
The mass of CaO that shall be obtained by heating 25 kg of \(60 \%\) pure lime stone is:
1 \(8.4 \times 10^3 \mathrm{gm}\)
2 \(1.80 \times 10^3 \mathrm{gm}\)
3 \(8.4 \times 10^{-3} \mathrm{gm}\)
4 \(8.96 \times 10^3 \mathrm{gm}\)
Explanation:
a
\[
\begin{aligned}
& \mathrm{CaCO}_3 \longrightarrow \\
& 100 \mathrm{~kg} \quad 56 \mathrm{~kg} \\
& \text { Pure limestone in } 25 \mathrm{~kg}=60 \% \text { of } 25=15 \mathrm{~kg} \\
& 100 \mathrm{~kg} \mathrm{CaCO}{ }_3 \text { give } \mathrm{CaO}=56 \mathrm{~kg} \\
& 15 \mathrm{~kg} \mathrm{CaCO}_3 \text { give } \mathrm{CaO}=\frac{56}{100} \times 15=8.4 \mathrm{~kg} \\
& =8.4 \times 10^3 \mathrm{gm}
\end{aligned}
\]
SECTION-B