Calculate the mass of lime that can be prepared by heating 200 kg of limestone that is 90% pure `CaCO_3`
`CaCO_3 to CaO + CO_2`
`100 kg xx 10^(-3)" " 56 kg xx 10^(-3)`

Explanation

`200kg ` of `90%` pure `CaCO_3 = 200 xx (90)/(100)`
`=180 kg ` pure ` CaCO_3`
` 100 xx 10^(-3)` kg of pure `CaCO_3 ` on heating gives ` 56 xx 10^(-3)` kg of CaO
180 kg of `CaCO_3` gives on heating `=(56 xx 10^(-3) xx 180)/( 100 xx 10^(-3))`
`= 100.8 kg ` CaO