MCQ
How much lime $(CaO)$ can be obtained by heating $200\, kg$ of limestone which is $95\%$ pure ? ............... $\mathrm{kg}$
- ✓$106.4$
- B$53.2$
- C$10.64$
- D$5.32$
molecular mass of $CaO=56$
$100 \mathrm{g}$ $CaCO_3$ make $56 \mathrm{g}$ $CaO$
$200000 \mathrm{g}$ will make $=56\times 200000 / 100$
$=112 \mathrm{kg}$
for $95 \%=106.40 \mathrm{kg}$ of $\mathrm{CaO}$
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[Atomic mass : $\mathrm{H}: 1.0, \mathrm{C}: 12.0, \mathrm{O}: 16.0]$
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