Đáp án + Giải thích các bước giải:

$n_{CuCl_2}=\dfrac{200.13,5\%}{135}=0,2(mol)$

$n_{AgNO_3}=\dfrac{300.17\%}{170}=0,3(mol)$

$a)CuCl_2+2AgNO_3\to Cu(NO_3)_2+2AgCl\downarrow$

Xét tỉ lệ: $n_{CuCl_2}>\dfrac{n_{AgNO_3}}{2}\to CuCl_2$ dư

Theo PT: $n_{AgCl}=n_{AgNO_3}=0,3(mol)$

$\to m_{AgCl}=0,3.143,5=43,05(g)$

$b)$ Theo PT: $n_{Cu(NO_3)_2}=0,5n_{AgNO_3}=0,15(mol)$

$\to m_{Cu(NO_3)_2}=0,15.188=28,2(g)$

Theo PT: $n_{CuCl_2(dư)}=0,2-0,5n_{AgNO_3}=0,05(mol)$

$\to m_{CuCl_2(dư)}=0,05.135=6,75(g)$

$m_{\text{dd sau}}=200+300-43,05=456,95(g)$

$\to C\%_{CuCl_2(dư)}=\dfrac{6,75}{456,95}.100\%\approx 1,48\%$

$\to C\%_{Cu(NO_3)_2}=\dfrac{28,2}{456,95}.100\%\approx 6,17\%$

a. PTHH: CuCl2+2AgNO3→2AgCl↓+Cu(NO3)2

mCuCl2=`(200.13,5)/(100)`=27g

→nCuCl2=`(27)/(135)`=0,2mol

mAgNO3=`(300.17)/(100)`=51g

→nAgNO3=`(51)/(170)`=0,3mol

Xét tỉ lệ: `(n_{CuCl_2})/1> (n_{AgNO_3})/2`

→CuCl2 dư, AgNO3 hết

Theo PTHH: nAgCl=nAgNO3=0,3mol

→a=mAgCl=0,3.143,5=43,05g

b. Theo PTHH: nCuCl2(pư)=nCu(NO3)2=`1/2`nAgNO3=0,15mol

mdd spư=200+300-43,05=456,95g

C% CuCl2(dư)=`((0,2-0,15).135)/(456,95).100`=1,477%

C% Cu(NO3)2=`(0,15.188)/(456,95).100`=6,17%