Đáp án:
\(C{\% _{CuS{O_4}}} = 4,6875\% \)
\(C{\% _{{H_2}S{O_4}}} = 2,793\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(CuO + {H_2}S{O_4}\xrightarrow{{}}CuS{O_4} + {H_2}O\)
Ta có:
\({n_{CuO}} = \frac{{12}}{{64 + 16}} = 0,15{\text{ mol}}\)
\({m_{{H_2}S{O_4}}} = 500.5,8\% = 29{\text{ gam}}\)
\( \to {n_{{H_2}S{O_4}}} = \frac{{29}}{{98}} > {n_{CuO}}\)
Vậy axit dư
\({n_{CuS{O_4}}} = {n_{CuO}} = 0,15{\text{ mol}}\)
\( \to {m_{CuS{O_4}}} = 0,15.160 = 24{\text{ gam}}\)
\({n_{{H_2}S{O_4}{\text{ phản ứng}}}} = {n_{CuO}} = 0,15{\text{ mol}}\)
\( \to {m_{{H_2}S{O_4}{\text{ phản ứng}}}} = 0,15.98 = 14,7{\text{ gam}} \to {{\text{m}}_{{H_2}S{O_4}{\text{ dư}}}} = 29 - 14,7 = 14,3{\text{ gam}}\)
BTKL:
\({m_{dd{\text{sau phản ứng}}}} = {m_{CuO}} + {m_{dd{{\text{H}}_2}S{O_4}}} = 12 + 500 = 512{\text{ gam}}\)
\( \to C{\% _{CuS{O_4}}} = \frac{{24}}{{512}} = 4,6875\% \)
\(C{\% _{{H_2}S{O_4}}} = \frac{{14,3}}{{512}} = 2,793\% \)
