Đáp án:
a, \({V_{S{O_2}}} = 1,96l\)
b, \(C{M_{{H_2}S{O_4}}} = \dfrac{{0,675}}{{0,5}} = 1,35M\)
Giải thích các bước giải:
\(\begin{array}{l}
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
{n_{Cu}} = 0,0875mol\\
\to {n_{S{O_2}}} = {n_{Cu}} = 0,0875mol\\
\to {V_{S{O_2}}} = 1,96l\\
{H_2}S{O_4} + 2NaOH \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{NaOH}} = \dfrac{{200 \times 20}}{{100 \times 40}} = 1mol\\
\to {n_{{H_2}S{O_4}}}dư= \dfrac{1}{2}{n_{NaOH}} = 0,5mol\\
\to {n_{{H_2}S{O_4}}}đầu= 2{n_{Cu}} + {n_{{H_2}S{O_4}}}dư= 0,675mol\\
\to C{M_{{H_2}S{O_4}}} = \dfrac{{0,675}}{{0,5}} = 1,35M
\end{array}\)
