a)
$CuCl_2 + 2AgNO_3 \to 2AgCl + Cu(NO_3)_2$
$n_{AgNO_3} = \dfrac{150.10\%}{170} = \dfrac{3}{34}(mol)$
Theo PTHH :
$n_{CuCl_2} = \dfrac{1}{2}n_{AgNO_3} = \dfrac{3}{68}(mol)$
$C\%_{CuCl_2} = \dfrac{ \dfrac{3}{68}.135}{150}.100\% = 3,97\%$
b)
Sau phản ứng :
$m_{dd} = m_{dd\ CuCl_2} + m_{dd\ AgNO_3} - m_{AgCl}$
$= 150 + 150 - \dfrac{3}{34}.143,5 = 287,33(gam)$
$C\%_{Cu(NO_3)_2} = \dfrac{ \dfrac{3}{68}.188}{287,33}.100\% = 2,88\%$