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Answer 1 · 2017-05-20T05:03:21

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%#