Question

Question #11d9c

Answer 1

Calcium carbonate `CaCO_3(s)` on heating strongly decomposes as follows

`CaCO_3(s)" " ->" "CaO(s)" "+" "CO_2(g)uarr`

Molar mass `(40+12+3xx16)" "(40+16)" "" "(12+2xx16)`

`=100g"/"mol" "" "=56g"/"mol" "=44g"/"mol`

From the above balanced equation we see that

56g solid residue `CaO(s)` is left from 100g `CaCO_3(s)`

So 2.2 g solid residue `CaO(s)` will come from `(100xx2.2)/56`g `CaCO_3(s)`

Hence 5g contaninated `CaCO_3(s)` contains `(100xx2.2)/56`g pure `CaCO_3(s)`

So percentage of purity of

`CaCO_3(s)=(100xx2.2)/56xx100/5%~~78.6%`