Question

10g sample of a mixture of CaCl_2 and NaCl is treated to precipitate all the calcium as CaCO_3.This CaCO_3 is heated to convert all the Ca to CaO and the final mass of CaO is 1.68g.The percent by mass of CaCl_2 in the original mixture is?

Answer 1

CaCO3 +heat--->CaO +Co2
X gms               1.68gms
100                   56
[molecular mass of CaCO3=100u
molecular mass of CaO=56u]
cross multiply through the equation and solve for x grams of caco3
x=1.62x100/56
=2.89g
Cacl2 +Na2Co3 --->Caco3 +2 Nacl
x                             2.89g
111                         100
Again cross multiply through the equation and solve for x grams of Cacl2
x=2.89x 111/100
=3.2g
%CaCl2 in original mixture 3.2grams
Cacl2 over 10gms mixture is 32%

Answer 2

mCaO=1.68xg

Mr[CaO]=56

∴nCaO=1.6856=0.03

Action of heat on CaCO3 gives:

CaCO3→CaO+CO2

∴nCaCO3=0.03

This was formed from:

Ca2+(aq)+CO2−3(aq)→CaCO3(s)

∴nCaCl2=0.03

Mr[CaCl2]=40+71=111

∴mCaCl2=0.03×111=3.33xg

∴ percentage CaCl2 =3.3310×100=33.3  Approx. .i.e 32 %