The mass of the mixture containing calcium carbonate and calcium hydroxide decreased after burning to 60% of the initial value. How to calculate the mass ratio of CaCO3 and Ca(OH)2 in the pre-burning mixture?

M1= M(Ca(OH))= 74 g/mol
M2= M(CaCO3)= 100 g/mol
M3= M(CaO)= 56 g/mol

1 Answer
P dilip_k
Feb 17, 2018

The balanced equations of the thermal decomposition reaction occurring here are

#CaCO_3(s)->CaO(s)+" "CO_2(g)uarr#
#" "1mol" "" "1mol" "" "" "1mol#

#" "100g" "" "56g" "" "" "44g#

#Ca(OH)_2(s)->"CaO(s)+" "H_2O(g)uarr#

#" "1mol" "" "" "1mol" "" "" "1mol"#

#" "74g" "" "" "" "56g" "" "" "18g#

Let the masses of CaCO3 and Ca(OH)2 in the pre-burning mixture are #xg and yg#
respectively.

Here loss in mass occurs due to removal of volatile matter #CO_2andH_2O#

So total loss in weight will be

#=44/100x+18/74y# g

Given that the mass of the mixture containing calcium carbonate and calcium hydroxide decreased after burning to 60% of the initial value. So loss in mass is #40%# of initial value.

Hence

#(x+y)*40/100=44/100x+18/74y#

#=>y*40/100-18/74y=4/100x#

#=>40*74y-18*100y=4*74x#

#=>1160y=296x#

#=>x/y=1160/296=145/37#

So the ratio of masses

#CaCO_3:Ca(OH)_2=145:37#

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