NaCl +AgNO3--->NaNO3 +AgCl(1)
x---------------------------------x
KCl +AgNO3--->KNO3 +AgCl(2)
y--------------------------------y
Ta có
n\(_{AgCl}=\frac{8,607}{143,5}=0,06\left(mol\right)\)
Suy ra ta có hệ pt
\(\left\{{}\begin{matrix}58,5x+74,5y=3,9\\x+y=0,06\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,036\\y=0,024\end{matrix}\right.\)
%m\(_{NaCl}=\frac{0,036.58,5}{3,9}.100\%=54\%\)
%m\(_{KCl}=100\%-54\%=46\%\)
Chúc bạn học tốt
Gọi số mol trong hh NaCl, KCl lần lượt là x, y ( mol)
\(PTHH:NaCl+AgNO3\rightarrow AgCl+NaN\text{O}3\left(1\right)\)\(KCl+AgNO3\rightarrow AgCl+KNO3\left(2\right)\)
\(n_{AgCl}=\frac{8,607}{143,5}=0,06\)
Từ (1) và (2) suy ra
\(\Rightarrow\left\{{}\begin{matrix}58,5x+74,5y=3,9\\x+y=0,06\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2356\\y=0,0243\end{matrix}\right.\)
\(\Rightarrow nNaCl=0,0356\Rightarrow mNaCl=2,0826\)
\(\Rightarrow mKCl=1,8134\)
\(\Rightarrow\%mNaCl=\frac{2,0826.100\%}{3,9}=53,4\%\)
\(\Rightarrow\%mKCl=46,6\%\)