The correct options are
A The % purity of Na2CO3 is 50%
C The number of moles of Na2CO3= that of CaCO3=0.1 mol
Molar mass of CaCO3=100 g mol−1
Molar mass of Na2CO3=106 g mol−1
Na2CO3+CaCl2→CaCO3+2NaCl
(a) From stoichiometry,
moles of CaCO3=10100=0.1 mol
moles of Na2CO3= moles of CaCO3=12× moles of NaCl
mass of Na2CO3=0.1×106=10.6 g
% purity Na2CO3=10.621.2=100=50%
c) moles of CaCO3=10100=0.1 mol
(d) Moles of NaCl=2×0.1=0.2 mol