Đáp án:
\(C{\% _{KCl}} = 13,85\% \)
Giải thích các bước giải:
\(\begin{array}{l}
2K + 2HCl \to 2KCl + {H_2}\\
{n_K} = \dfrac{{29,25}}{{39}} = 0,75mol\\
{m_{HCl}} = \dfrac{{250 \times 7,3}}{{100}} = 18,25g\\
{n_{HCl}} = \dfrac{{18,25}}{{36,5}} = 0,5mol\\
\dfrac{{0,75}}{1} > \dfrac{{0,5}}{1} \Rightarrow K\text{ dư}\\
{n_{{K_{pu}}}} = {n_{HCl}} = 0,5mol\\
{n_{{H_2}}} = \dfrac{{{n_{HCl}}}}{2} = 0,25mol\\
{n_{KCl}} = {n_{HCl}} = 0,5mol\\
{m_{KCl}} = 0,5 \times 74,5 = 37,25g\\
{m_{{\rm{dd}}spu}} = 0,5 \times 39 + 250 - 0,25 \times 2 = 269g\\
C{\% _{KCl}} = \dfrac{{37,25}}{{269}} \times 100\% = 13,85\%
\end{array}\)
