Question

Calculate the weight of lime ($CaO$) that can be obtained by heating 200 kg of limestone which is 93% pure. How many moles of impure potassium chlorate of 75% purity is required to produce 48 g of oxygen?

Answer 1

$CaC{O}_3\to CaO+C{O}_2$

moles of $CaC{O}_3$ in 200 kg = 1000 moles

moles of CaO produced = 1000 moles

weight of CaO produced = 1000* 56 = 56000g

limestone is 90% pure

therefore the amount of CaO produced = $56000\ast \frac{93}{100}$ = 52080g or 52.08 kg.

we have,

$KCl{O}_3\to KCl+3/2O_2$

moles of $O_2$ in 48 g = $\frac{48}{32}$ = 1.5

that means 1 mole of $KCl{O}_3$ is dissociating.

1 mole of $KCl{O}_3$ = 122.5 g

so, amount of impure $KCl{O}_3$ = $122.5\ast \frac{100}{75}$ = 163.3 g