Đáp án:
\(\begin{array}{l}
{C_\% }HCl = 2,48\% \\
{C_\% }KCl = 3,04\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
KOH + HCl \to KCl + {H_2}O\\
{n_{KOH}} = \dfrac{{168 \times 5\% }}{{56}} = 0,15\,mol\\
{n_{HCl}} = \dfrac{{200 \times 7,3\% }}{{36,5}} = 0,4\,mol\\
{n_{KOH}} < {n_{HCl}} \Rightarrow HCl \text{ dư }\\
{n_{HCl}} \text{ dư }= 0,4 - 0,15 = 0,25\,mol\\
{n_{KCl}} = {n_{KOH}} = 0,15\,mol\\
{m_{{\rm{dd}}spu}} = 168 + 200 = 368g\\
{C_\% }HCl \text{ dư }= \dfrac{{0,25 \times 36,5}}{{368}} \times 100\% = 2,48\% \\
{C_\% }KCl = \dfrac{{0,15 \times 74,5}}{{368}} \times 100\% = 3,04\%
\end{array}\)
