Đáp án:
\(\begin{array}{l}
a/\\
{m_{ZnS{O_4}}} = 24,15g\\
{m_{CuS{O_4}}} = 1,44g\\
b/\\
{m_{ZnC{l_2}}} = 20,4g\\
{m_{CuC{l_2}}} = 1,215g\\
c/{V_{S{O_2}}} = 3,36l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
5.\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
{n_{{H_2}}} = 0,15mol\\
\to {n_{Zn}} = {n_{{H_2}}} = 0,15mol\\
\to {m_{Zn}} = 9,75g\\
\to {m_{CuO}} = 0,75g \to {n_{CuO}} = 0,009mol
\end{array}\)
\(\begin{array}{l}
a/\\
{n_{ZnS{O_4}}} = {n_{Zn}} = 0,15mol\\
{n_{CuS{O_4}}} = {n_{CuO}} = 0,009mol\\
\to {m_{ZnS{O_4}}} = 24,15g\\
\to {m_{CuS{O_4}}} = 1,44g
\end{array}\)
\(\begin{array}{l}
b/\\
{n_{{H_2}S{O_4}}} = 0,2mol\\
{n_{{H_2}S{O_4}(pt)}} = {n_{Zn}} + {n_{CuO}} = 0,16mol\\
\to {n_{{H_2}S{O_4}(dư)}} = 0,04mol
\end{array}\)
\(\begin{array}{l}
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl\\
BaC{l_2} + ZnS{O_4} \to BaS{O_4} + ZnC{l_2}\\
BaC{l_2} + CuS{O_4} \to BaS{O_4} + CuC{l_2}\\
{n_{ZnC{l_2}}} = {n_{ZnS{O_4}}} = 0,15mol\\
{n_{CuC{l_2}}} = {n_{CuS{O_4}}} = 0,009mol\\
\to {m_{ZnC{l_2}}} = 20,4g\\
\to {m_{CuC{l_2}}} = 1,215g
\end{array}\)
\(\begin{array}{l}
c/\\
Zn + 2{H_2}S{O_4} \to ZnS{O_4} + S{O_2} + 2{H_2}O\\
{n_{S{O_2}}} = {n_{Zn}} = 0,15mol\\
\to {V_{S{O_2}}} = 3,36l
\end{array}\)
