Đặt \(\left\{{}\begin{matrix}n_{KCl}=a\left(mol\right)\\n_{KBr}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
\(KCl+AgNO_3\rightarrow AgCl\downarrow+KNO_3\)
a------->a----------->a
\(KBr+AgNO_3\rightarrow AgBr\downarrow+KNO_3\)
b------->b----------->b
Do \(m_{\downarrow}=m_{AgNO_3}\)
=> 170a + 170b = 143,5a + 188b
=> 26,5a = 18b
=> \(b=\dfrac{53}{36}a\)
=> \(\%m_{KCl}=\dfrac{74,5a}{74,5a+119b}.100\%=\dfrac{74,5a}{74,5a+119.\dfrac{53}{36}a}.100\%=29,84\%\)
=> \(\%m_{KBr}=100\%-29,84\%=70,16\%\)