hh NaCl NaNO3 dd D NaNO3, KNO3, Mg(NO3)2 + NaOH $\rightarrow$ Mg(OH)2 $\rightarrow$ MgO
KCl + AgNO3 $\rightarrow$KNO3 +Mg $\rightarrow$
MgCl2 Mg(NO3)2 tủa C :Ag
+ HCl $\rightarrow$ MgCl2 , Ag
Mg dư
m Mg pư=m tủa C giảm= 1,844 (g)
=> nMg pư= $\frac{2-1,844}{24}$ = 0,0065 (mol)
Mg + 2AgNO3 $\rightarrow$ Mg(NO3)2 + 2Ag
=>nAgNO3 dư=2 nMg pư= 0,013 (mol)
=> nAgNO3 pư= 0,12-0,013 = 0,107= nAgCl => mtủa A=0,107 . 143,5 = 15,3545 (g)
nAg=nAgNO3 dư= 0,013 mol => m tủa C=3,248 (g)
n Mg pư= 0,0065 mà n Mg(trong MgO)= 0,3/40 = 0,0075 => nMgCl2 = 0,0075- 0, 0065 = 0,001 => % MgCl2= 1,504%
=> mNaCl,KCl = 6,3175- 0,001.95=6,2225 (g)
=> n AgNO3 pư vs MgCl2= 2 nMgCl2 = 0,002 mol
=> nAgNO3 pư vs NaCl, KCl = 0,107- 0,002=0,105 mol
giải hệ 58,5 x + 74,5y= 6,2225
x + y = 0,105
=> x= 0,1 mol ; y=0, 005
=> % NaCl = 92,6% % KCl = 5,896%