ptpu: $NaCl+AgNO_3\rightarrow AgCl+NaNO_3$
$x$ $x$
$KCl+AgNO_3\rightarrow AgCl+KNO_3$
$y$ $y$
theo bai ra ta co : $n _{AgCl}=\frac{57,4}{143,5}=0,4$ mol
$\Rightarrow x+y=0,4(1)$
lai co: $58,5x+74,5y=26,6(2)$
tu $(1),(2)\Rightarrow \left\{ \begin{array}{l} x=0,2\\ y=0,2 \end{array} \right.$
$m_{NaCl}=58,5.0,2=11,6g$
$m_{KCl}=74,5.0,2=14,9g$
cong thuc:$C$%=$\frac{m_{ct}}{m_{dd}}.100$%
ban tu tinh nha,muon oy lai k co mt ...:)