Đáp án:
\(\begin{array}{l}
a)\\
A:{H_2},{H_2}S\\
b)\\
{m_{Zn}} = 13g\\
c)\\
{V_{{H_2}S}} = 2,24l\\
d)\\
{V_{{\rm{dd}}CuS{O_4}}} = 0,2l\\
e)\\
{m_{{H_2}O}} = 5,4g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
FeS + 2HCl \to FeC{l_2} + {H_2}S\\
A:{H_2},{H_2}S\\
b)\\
{H_2}S + CuS{O_4} \to CuS + {H_2}S{O_4}\\
{n_{CuS}} = \dfrac{{9,6}}{{96}} = 0,1\,mol\\
{n_A} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{{H_2}S}} = {n_{CuS}} = 0,1\,mol\\
{n_{{H_2}}} = 0,3 - 0,1 = 0,2\,mol\\
{n_{Zn}} = {n_{{H_2}}} = 0,2\,mol\\
{m_{Zn}} = 0,2 \times 65 = 13g\\
c)\\
{V_{{H_2}S}} = 0,1 \times 22,4 = 2,24l\\
d)\\
{n_{CuS{O_4}}} = {n_{{H_2}S}} = 0,1\,mol\\
{V_{{\rm{dd}}CuS{O_4}}} = \frac{{0,1}}{{0,5}} = 0,2l\\
e)\\
2{H_2} + {O_2} \xrightarrow{t^0} 2{H_2}O\\
2{H_2}S + 3{O_2} \xrightarrow{t^0} 2S{O_2} + 2{H_2}O\\
{n_{{H_2}O}} = 0,2 + 0,1 = 0,3\,mol\\
{m_{{H_2}O}} = 0,3 \times 18 = 5,4g
\end{array}\)
