Đáp án:
\(\begin{array}{l}
a.{m_{{H_2}S{O_4}(dư)}} = 19,6g\\
b.\\
C{\% _{{H_2}S{O_4}(dư)}} = 12,16\% \\
C{\% _{MgS{O_4}}} = 14,89\% \\
c.H = 125\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
{n_{Mg}} = 0,2mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{156,8 \times 25\% }}{{100\% }} = 39,2g\\
\to {n_{{H_2}S{O_4}}} = 0,4mol\\
\to {n_{{H_2}S{O_4}}} > {n_{Mg}} \to {n_{{H_2}S{O_4}}}dư
\end{array}\)
\(\begin{array}{l}
a.\\
{n_{{H_2}S{O_4}}} = {n_{Mg}} = 0,2mol\\
\to {n_{{H_2}S{O_4}(dư)}} = 0,2mol\\
\to {m_{{H_2}S{O_4}(dư)}} = 19,6g
\end{array}\)
\(\begin{array}{l}
b.\\
{n_{MgS{O_4}}} = {n_{{H_2}}} = {n_{Mg}} = 0,2mol\\
\to {m_{MgS{O_4}}} = 24g\\
\to {m_{{H_2}}} = 0,4g\\
{m_{{\rm{dd}}}} = {m_{Mg}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{{H_2}}} = 161,2g\\
\to C{\% _{{H_2}S{O_4}(dư)}} = \dfrac{{19,6}}{{161,2}} \times 100\% = 12,16\% \\
\to C{\% _{MgS{O_4}}} = \dfrac{{24}}{{161,2}} \times 100\% = 14,89\%
\end{array}\)
\(\begin{array}{l}
c.\\
CuO + {H_2} \to Cu + {H_2}O\\
{n_{Cu}} = 0,25mol\\
\to {n_{Cu}} > {n_{{H_2}}} \to {n_{Cu}}dư\\
\to {n_{Cu}} = {n_{{H_2}}} = 0,2mol\\
\to {m_{Cu}} = 12,8g\\
\to H = \dfrac{{16 \times 100}}{{12,8}} = 125\%
\end{array}\)
