Đáp án:

Bạn tham khảo lời giải ở dưới nha ! 

Giải thích các bước giải:

\(\begin{array}{l}
2.\\
K + HCl \to KCl + \dfrac{1}{2}{H_2}\\
{n_K} = 0,4mol\\
{m_{HCl}} = \dfrac{{120 \times 7,3\% }}{{100\% }} = 8,76g\\
 \to {n_{HCl}} = 0,24mol\\
 \to {n_{HCl}} < {n_K} \to {n_K}dư
\end{array}\)

 \(\begin{array}{l}
{n_{KCl}} = {n_K} = {n_{HCl}} = 0,24mol\\
 \to {n_{K(dư)}} = 0,16mol \to {m_{K(dư)}} = 6,24g\\
 \to {m_{KCl}} = 17,88g\\
{n_{{H_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,12mol \to {m_{{H_2}}} = 0,24g\\
 \to {m_{{\rm{dd}}}} = {m_K} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 135,36g\\
 \to C{\% _{KCl}} = \dfrac{{17,88}}{{135,36}} \times 100\%  = 19,2\% \\
 \to C{\% _{K(dư)}} = \dfrac{{6,24}}{{135,36}} \times 100\%  = 4,6\% 
\end{array}\)

\(\begin{array}{l}
3.\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
{n_{Zn}} = 0,2mol\\
 \to {n_{{H_2}S{O_4}}} = {n_{Zn}} = 0,2mol\\
 \to {m_{{H_2}S{O_4}}} = 19,6g
\end{array}\)

\(\begin{array}{l}
a)\\
x = C{\% _{{H_2}S{O_4}}} = \dfrac{{19,6}}{{150}} \times 100\%  = 13,07\% 
\end{array}\)

\(\begin{array}{l}
b)\\
{n_{ZnS{O_4}}} = {n_{{H_2}}} = {n_{Zn}} = 0,2mol\\
 \to {m_{ZnS{O_4}}} = 32,2g\\
{m_{{\rm{dd}}}} = {m_{Zn}} + {m_{{H_2}S{O_4}}} - {m_{{H_2}}} = 162,6g\\
 \to C{\% _{ZnS{O_4}}} = \dfrac{{32,2}}{{162,6}} \times 100\%  = 19,8\% 
\end{array}\)