PTHH: \({H_2}S + CuS{O_4}\xrightarrow{{}}CuS \downarrow + {H_2}S{O_4}\)
\({n_{CuS}} = \dfrac{{1,92}}{{96}} = 0,02\left( {mol} \right) \Rightarrow {n_{CuS{O_4}}} = {n_{CuS}} = 0,02\left( {mol} \right)\)
\( \Rightarrow C{\% _{CuS{O_4}}} = \dfrac{{0,02.160}}{{16}}.100\% = 20\% \)
=> Chọn A
