Đáp án:
\( C{\% _{MgS{O_4}}} = 9,677\% \)
\( C{\% _{{H_2}S{O_4}}} = 1,58\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(MgO + {H_2}S{O_4}\xrightarrow{{}}MgS{O_4} + {H_2}O\)
Ta có:
\({n_{MgO}} = \frac{{20}}{{24 + 16}} = 0,5{\text{ mol}}\)
\( \to {m_{{H_2}S{O_4}}} = 600.9,8\% = 58,8{\text{ gam}}\)
\( \to {n_{{H_2}S{O_4}}} = \frac{{58,8}}{{98}} = 0,6{\text{ mol > }}{{\text{n}}_{MgO}}\)
Do vậy axit dư
BTKL:
\({m_{dd}} = {m_{MgO}} + {m_{dd{{\text{H}}_2}S{O_4}}} = 20 + 600 = 620{\text{ gam}}\)
\({n_{MgS{O_4}}} = {n_{MgO}} = 0,5{\text{ mol}}\)
\({n_{{H_2}S{O_4}{\text{ dư}}}} = 0,6 - 0,5 = 0,1{\text{ mol}}\)
\( \to {m_{MgS{O_4}}} = 0,5.(24 + 96) = 60{\text{ gam}}\)
\({m_{{H_2}S{O_4}{\text{ dư}}}} = 0,1.98 = 9,8{\text{ gam}}\)
\( \to C{\% _{MgS{O_4}}} = \frac{{60}}{{620}} = 9,677\% \)
\( \to C{\% _{{H_2}S{O_4}}} = \frac{{9,8}}{{620}} = 1,58\% \)
