Đáp án:
\(C{\% _{CuS{O_4}}} = 3,15\% \)
\(C{\% _{{H_2}S{O_4}}} = 17,76\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(CuO + {H_2}S{O_4}\xrightarrow{{}}CuS{O_4} + {H_2}O\)
Bảo toàn khối lượng:
\({m_{dd}} = {m_{CuO}} + {m_{dd\;{{\text{H}}_2}S{O_4}}} = 1,6 + 100 = 101,6{\text{ gam}}\)
\({n_{CuO}} = \frac{{1,6}}{{64 + 16}} = 0,02{\text{ mol}}\)
\({m_{{H_2}S{O_4}}} = 100.20\% = 20{\text{ gam}}\)
\( \to {n_{{H_2}S{O_4}}} = \frac{{20}}{{98}} > {n_{CuO}}\)
Vậy axit dư
\( \to {n_{CuS{O_4}}} = {n_{{H_2}S{O_4}{\text{ phản ứng}}}} = {n_{CuO}} = 0,02{\text{ mol}}\)
\( \to {m_{CuS{O_4}}} = 0,02.(64 + 96) = 3,2{\text{ gam}}\)
\({m_{{H_2}S{O_4}}} = 20 - 0,02.98 = 18,04{\text{ gam}}\)
\( \to C{\% _{CuS{O_4}}} = \frac{{3,2}}{{101,6}}.100\% = 3,15\% \)
\(C{\% _{{H_2}S{O_4}}} = \frac{{18,04}}{{101,6}}.100\% = 17,76\% \)
