Đáp án:
\(\begin{array}{l}
16.\\
a)\\
{m_{ZnS{O_4}}} = 22,54g\\
{V_{{H_2}}} = 3,136l\\
b){m_{{\rm{dd}}{H_2}S{O_4}}} = 80g\\
c)C{\% _{ZnS{O_4}}} = 25,38\%
\end{array}\)
\(\begin{array}{l}
17.\\
a)\\
{m_{MgS{O_4}}} = 14,4g\\
{V_{{H_2}}} = 2,688l\\
b){m_{{\rm{dd}}{H_2}S{O_4}}} = 100g\\
c)C{\% _{MgS{O_4}}} = 14\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
16.\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
{n_{Zn}} = 0,14mol\\
a)\\
{n_{ZnS{O_4}}} = {n_{{H_2}}} = {n_{Zn}} = 0,14mol\\
\to {m_{ZnS{O_4}}} = 22,54g\\
\to {V_{{H_2}}} = 3,136l
\end{array}\)
\(\begin{array}{l}
b)\\
{n_{{H_2}S{O_4}}} = {n_{Zn}} = 0,14mol\\
\to {m_{{H_2}S{O_4}}} = 13,72g\\
\to {m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{13,72}}{{17,15\% }} \times 100\% = 80g
\end{array}\)
\(\begin{array}{l}
c)\\
{m_{{\rm{dd}}}} = {m_{Zn}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{{H_2}}} = 88,82g\\
\to C{\% _{ZnS{O_4}}} = \dfrac{{22,54}}{{88,82}} \times 100\% = 25,38\%
\end{array}\)
\(\begin{array}{l}
17.\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
{n_{Mg}} = 0,12mol
\end{array}\)
\(\begin{array}{l}
a)\\
{n_{MgS{O_4}}} = {n_{{H_2}}} = {n_{Mg}} = 0,12mol\\
\to {m_{MgS{O_4}}} = 14,4g\\
\to {V_{{H_2}}} = 2,688l
\end{array}\)
\(\begin{array}{l}
b)\\
{n_{{H_2}S{O_4}}} = {n_{Mg}} = 0,12mol\\
\to {m_{{H_2}S{O_4}}} = 11,76g\\
\to {m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{11,76}}{{11,76\% }} \times 100\% = 100g
\end{array}\)
\(\begin{array}{l}
c)\\
{m_{{\rm{dd}}}} = {m_{Mg}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{{H_2}}} = 102,64g\\
\to C{\% _{MgS{O_4}}} = \dfrac{{14,4}}{{102,64}} \times 100\% = 14\%
\end{array}\)
