`a)\ Cu+2AgNO_3->Cu(NO_3)_2+2Ag`

`b)`

Đặt `n_{Cu}=x(mol)`

Theo PT: `n_{Ag}=2x(mol)`

Tăng giảm khối lượng:

`m_{Ag}-m_{Cu}=\Delta m=1,52(g)`

`->2x.108-64x=1,52`

`->x=0,01(mol)`

Theo PT: `n_{AgNO_3}=2x=0,02(mol)`

`200ml=0,2l`

`->C_{M\ AgNO_3}={0,02}/{0,2}=0,1M`

`c)`

Theo PT: `n_{Cu(NO_3)_2}=x=0,01(mol)`

`->C\%_{Cu(NO_3)_2}={0,01.188}/{200.1,1}.100\%\approx 0,85\%`

Đáp án:

$C_{M\ dd\ AgNO_3}=0,1(M)\\ C\%_{dd\ Cu(NO_3)_2}=0,85\%$

Giải thích các bước giải:

$a,\\ Cu+2AgNO_3\to Cu(NO_3)_2+2Ag$

$b,\\ Đặt\ n_{Cu\ pư}=x(mol)\\ n_{Ag}=2.n_{Cu}=2x(mol)\\ \Rightarrow m_{Ag}-m_{Cu}=1,52(g)\\ \Rightarrow 2x.108-64x=1,52\\ \Rightarrow x=0,01\\ n_{AgNO_3}=2.n_{Cu}=0,02(mol)\\ C_{M\ dd\ AgNO_3}=\dfrac{0,02}{0,2}=0,1(M)$

$c,\\ m_{dd\ Cu(NO_3)_2}=200.1,1=220(g)\\ n_{Cu(NO_3)_2}=n_{Cu}=0,01(mol)\\ C\%_{dd\ Cu(NO_3)_2}=\dfrac{0,01.188}{220}.100\%=0,85\%$