Đáp án:
\( {m_{CaC{O_3}}} = 5{\text{ gam}}; {m_{NaCl}} = 3,8{\text{ gam}}\)
\( {V_{dd{\text{rượu}}}} = 12,28633{\text{ ml}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(CaC{O_3} + 2C{H_3}COOH\xrightarrow{{}}{(C{H_3}COO)_2}Ca + C{O_2} + {H_2}O\)
Ta có:
\({n_{C{O_2}}} = \frac{{1,12}}{{22,4}} = 0,05{\text{ mol = }}{{\text{n}}_{CaC{O_3}}}\)
\( \to {m_{CaC{O_3}}} = 0,05.100 = 5{\text{ gam}} \to {m_{NaCl}} = 3,8{\text{ gam}}\)
\({C_2}{H_5}OH + {O_2}\xrightarrow{{men}}C{H_3}COOH + {H_2}O\)
Ta có:
\({m_{C{H_3}COOH}} = 2{n_{C{O_2}}} = 0,1{\text{ mol}}\)
\( \to {n_{{C_2}{H_5}OH{\text{ lt}}}} = {n_{C{H_3}COOH}} = 0,1{\text{ mol}}\)
Vì hiệu suất là \(80\%\)
\( \to {n_{{C_2}{H_5}OH}} = \frac{{0,1}}{{80\% }} = 0,125{\text{ mol}}\)
\( \to {m_{{C_2}{H_5}OH}} = 0,125.46 = 5,75{\text{ gam}}\)
\( \to {V_{{C_2}{H_5}OH}} = \frac{{5,75}}{{0,78}} = 7,3718{\text{ ml}}\)
\( \to {V_{dd{\text{rượu}}}} = \frac{{7,3718}}{{60\% }} = 12,28633{\text{ ml}}\)