nAgNO3 = 1.5*0.3=0.45(mol)
vì cho 2.4 g Mg vào A thấy có rắn C => AgNO3 dư
nMg(pư 4) =(2.4-1.92)/24=0.02(mol)
KCl + AgNO3 --> KNO3 + AgCl
a................a
MgCl2 + 2AgNO3 ---> Mg(NO3)2 + 2AgCl
0.08----------0.16-------( 0.1-0.02)
NaCl + AgNO3 ---> NaNO3 + AgCl
b...............b
Mg + 2AgNO3 ----> Mg(NO3)2 +2 Ag (4)
0.02---------0.04 ------------ 0.02 
dd D gồm NaNO3, KNO3 và Mg(NO3)2
NaOH + Mg(NO3)2---> Mg(OH)2 +NaNO3
-------------- 0.1 ------------ 0.1
nMgO= 4/40 =0.1(mol)
Mg(OH)2 ----->MgO + H2O
0.1 --------------0.1
n AgNO3 =0.45 = 0.04 +a+b+0.16
=> a+b=0.25(mol)
mMgCl2=0.08*95=7.6 g
Ta có hệ
a+b=0.25
74.5*a + 58.5*b=24.625-7.6
=> a=0.15 , b=0.1
%KCl =(74.5*0.15*100)/24.625=45.4%
%NaCl=(58.5*0.1*100)/24.625=23.8%
%MgCl2=100%-45.4%-23.8% =30.8%