Đáp án:
\(\begin{array}{l}
a)\\
{m_{Cu\,S{O_4}}} = 32g\\
b)\\
{V_{S{O_2}}} = 4,48l\\
c)\\
{m_{{H_2}S{O_4}}} = 39,2g\\
d)\\
{m_{N{a_2}S{O_3}}} = 25,2g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2{H_2}S{O_4} + Cu \to CuS{O_4} + S{O_2} + 2{H_2}O\\
a)\\
{n_{Cu}} = \dfrac{m}{M} = \dfrac{{12,8}}{{64}} = 0,2mol\\
{n_{CuS{O_4}}} = {n_{Cu}} = 0,2mol\\
{m_{Cu\,S{O_4}}} = n \times M = 0,2 \times 160 = 32g\\
b)\\
{n_{S{O_2}}} = {n_{Cu}} = 0,2mol\\
\Rightarrow {V_{S{O_2}}} = n \times 22,4 = 0,2 \times 22,4 = 4,48l\\
c)\\
{n_{{H_2}S{O_4}}} = 2{n_{Cu}} = 0,4mol\\
{m_{{H_2}S{O_4}}} = n \times M = 0,4 \times 98 = 39,2g\\
d)\\
S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,2mol\\
{m_{N{a_2}S{O_3}}} = n \times M = 0,2 \times 126 = 25,2g
\end{array}\)