Đáp án:
\( \ V {\text{ = 11}}{\text{,979 c}}{{\text{m}}^3}\)
\( m = 78,8{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\({C_2}{H_5}OH + 3{O_2}\xrightarrow{{{t^o}}}2C{O_2} + 3{H_2}O\)
\(Ba{(OH)_2} + C{O_2}\xrightarrow{{}}BaC{O_3} + {H_2}O\)
Ta có:
\({n_{kk}} = \frac{{67,2}}{{22,4}} = 3{\text{ mol}}\)
\( \to {n_{{O_2}}} = \frac{1}{5}{n_{kk}} = 0,6{\text{ mol}}\)
\( \to {n_{{C_2}{H_5}OH}} = \frac{1}{3}{n_{{C_2}{H_5}OH}} = 0,2{\text{ mol}}\)
\( \to {n_{C{O_2}}} = 2{n_{{C_2}{H_5}OH}} = 0,4{\text{ mol}}\)
\( \to {n_{BaC{O_3}}} = {n_{C{O_2}}} = 0,4{\text{ mol}}\)
\( \to m = {m_{BaC{O_3}}} = 0,4.197 = 78,8{\text{ gam}}\)
\({m_{{C_2}{H_5}OH}} = 0,2.46 = 9,2{\text{ gam}}\)
\( \to {V_{{C_2}{H_5}OH}} = \frac{{9,2}}{{0,8}} = 11,5{\text{ ml}}\)
\( \to V = \frac{{11,5}}{{96\% }} = 11,979{\text{ ml = 11}}{\text{,979 c}}{{\text{m}}^3}\)