Question

Calculate the mass of $\left(CaO\right)$ that can be prepared by heating 200 kg of limestone $CaC{O}_3$ which is 95% pure.

Answer 1

Amount of pure $CaC{O}_3=\frac{95}{100}\times 200=190kg$ $=190000g$
$CaC{O}_3\left(s\right)\to CaO\left(s\right)+C{O}_2\left(g\right)$

1 mole $CaC{O}_3\equiv$ 1 mole $CaO$
100 g $CaC{O}_3\equiv$ 56 g $CaO$
$\because$ 100 g $CaC{O}_3$ give $56gCaO$
$\therefore$ 190000 g $CaC{O}_3$ will give $\frac{56}{100}\times 190000g=106400g=106.4kg$