Đáp án:
\({m_{muối}} = 16,4{\text{ gam}}\)
\(CH_3COOH\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2RCOOH + 2Na\xrightarrow{{}}2RCOONa + {H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}}\)
BTKL:
\({m_{RCOOH}} + {m_{Na}} = {m_{muối}} + {m_{{H_2}}}\)
\( \to 12 + 4,6 = {m_{muối}} + 0,1.2\)
\( \to {m_{muối}} = 16,4{\text{ gam}}\)
\({n_{RCOOH}} = 2{n_{{H_2}}} = 0,2{\text{ mol}}\)
\( \to {M_{RCOOH}} = {M_R} + 45 = \frac{{12}}{{0,2}} = 60\)
\(M_R=15 \to R:CH_3-\)
Vậy axit là \(CH_3COOH\)
\(2CH_3COOH + 2Na\xrightarrow{{}}2CH_3COONa + {H_2}\)