Mỗi phần đặt n_{MgCl_2}=x(mol);n_{CuCl_2}=y(mol)

Phần 1:

MgCl_2+2AgNO_3->Mg(NO_3)_2+2AgCl

CuCl_2+2AgNO_3->Cu(NO_3)_2+2AgCl

Theo PT: n_{AgCl}=2x+2y={28,7}/{143,5}=0,2(1)

Phần 2:

MgCl_2+2NaOH->Mg(OH)_2+2NaCl

CuCl_2+2NaOH->Cu(OH)_2+NaCl

Mg(OH)_2  $$\xrightarrow{t^o}$$  MgO+H_2O

Cu(OH)_2  $$\xrightarrow{t^o}$$  CuO+H_2O

Theo PT: n_{MgO}=n_{MgCl_2}=x(mol);n_{CuO}=n_{CuCl_2}=y(mol)

->40x+80y=6(2)

(1)(2)->x=y=0,05(mol)

-> Ban đầu n_{MgCl_2}=n_{CuCl_2}=0,05.2=0,1(mol)

->C\%_{CuCl_2}={0,1.135}/{100}.100\%=13,5\%

       C\%_{MgCl_2}={0,1.95}/{100}.100\%=9,5\%

Gọi n_{MgCl_2} = a(mol); n_{CuCl_2} = b(mol0

Phần I:

n_{AgCl} = (28,7)/(143,5) = 0,2(mol)

MgCl_2 + 2AgNO_3 -> 2AgCl + Mg(NO_3)_2

a -> 2a -> 2a

CuCl_2 + 2AgNO_3 -> 2AgCl +Cu(NO_3)_2

b -> 2b -> 2b

=> 2a + 2b = 0,2(1)

Phần II:

MgCl_2 + 2NaOH -> Mg(OH)_2 + 2NaCl

a -> 2a -> a

CuCl_2 + 2NaOH -> Cu(OH)_2 + 2NaCl

b -> 2b -> b

$$Mg(OH)_2 \xrightarrow{t^o} MgO + H_2O$$

a -> a

$$Cu(OH)_2 \xrightarrow{t^o} CuO + H_2O$$

b -> b

=> 40a + 80b = 6

=> a = b = 0,05

Vậy dd ban đầu có n_{MgCl_2} = n_{CuCl_2} = 0,05.2 = 0,1(mol)

C\%_{MgCl_2} = (0,1.95)/(100) .100\% = 9,5\%

C\%_{CuCl_2} = (0,1.135)/(100) .100\% = 13,5\%