Đáp án:
\(\begin{array}{l}
{C_\% }KCl = 1,49\% \\
{C_\% }CaC{l_2} = 4,44\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
KCl + AgN{O_3} \to AgCl + KN{O_3}\\
CaC{l_2} + 2AgN{O_3} \to Ca{(N{O_3})_2} + 2AgCl\\
{n_{AgCl}} = \dfrac{{71,75}}{{143,5}} = 0,5\,mol\\
hh:KCl(a\,mol),CaC{l_2}(b\,mol)\\
\left\{ \begin{array}{l}
74,5a + 111b = 29,65\\
a + 2b = 0,5
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,2\\
{C_\% }KCl = \dfrac{{0,1 \times 74,5}}{{500}} \times 100\% = 1,49\% \\
{C_\% }CaC{l_2} = \dfrac{{0,2 \times 111}}{{500}} \times 100\% = 4,44\%
\end{array}\)