Đáp án:
\({m_{{C_2}{H_5}OH}} = 9,2{\text{ gam}}\)
\( V = 2,24{\text{ lít}}\)
\({m_{{C_2}{H_5}ONa}} = 12,4{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2{C_2}{H_5}OH + 2Na\xrightarrow{{}}2{C_2}{H_5}ONa + {H_2}\)
Ta có:
\({n_{Na}} = \frac{{4,6}}{{23}} = 0,2{\text{ mol = }}{{\text{n}}_{{C_2}{H_5}OH}} = {n_{{C_2}{H_5}ONa}}\)
\( \to {m_{{C_2}{H_5}OH}} = 0,2.46 = 9,2{\text{ gam}}\)
\( \to {n_{{H_2}}} = \frac{1}{2}{n_{Na}} = 0,1{\text{ mol}}\)
\( \to V = {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
\({m_{{C_2}{H_5}ONa}} = 0,2.62 = 12,4{\text{ gam}}\)