Đáp án:
\(\begin{array}{l}
a/{m_{{H_2}S{O_4}(dư)}} = 0,49g\\
c/{m_{KOH}} = 2,24g\\
d/{m_{ZnS{O_4}}} = 2,415g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
a/\\
{n_{Zn}} = 0,015mol\\
{n_{{H_2}S{O_4}}} = 0,02mol\\
\to {n_{Zn}} < {n_{{H_2}S{O_4}}} \to {n_{{H_2}S{O_4}}}dư\\
\to {n_{{H_2}S{O_4}}} = {n_{Zn}} = 0,015mol\\
\to {n_{{H_2}S{O_4}(dư)}} = 0,005mol\\
\to {m_{{H_2}S{O_4}(dư)}} = 0,49g
\end{array}\)
\(\begin{array}{l}
c/\\
2KOH + ZnS{O_4} \to {K_2}S{O_4} + Zn{(OH)_2}\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{ZnS{O_4}}} = {n_{Zn}} = 0,015mol\\
{n_{KOH}} = 2{n_{ZnS{O_4}}} + 2{n_{{H_2}S{O_4}}} = 0,04mol\\
\to {m_{KOH}} = 2,24g
\end{array}\)
\(\begin{array}{l}
d/\\
{m_{ZnS{O_4}}} = 2,415g
\end{array}\)
