Đáp án:
\({V_{{H_2}}} = 2,24{\text{ lít}}\)
\({m_{dd{\text{ C}}{{\text{H}}_3}COOH}} = 60{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2C{H_3}COOH + Mg\xrightarrow{{}}{(C{H_3}COO)_2}Mg + {H_2}\)
Ta có:
\({n_{Mg}} = \frac{{2,4}}{{24}} = 0,1{\text{ mol}}\)
\( \to {n_{{H_2}}} = {n_{Mg}} = 0,1{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
\({n_{C{H_3}COOH}} = 2{n_{{H_2}}} = 0,1.2 = 0,2{\text{ mol}}\)
\( \to {m_{C{H_3}COOH}} = 0,2.60 = 12{\text{ gam}}\)
\( \to {m_{dd{\text{ C}}{{\text{H}}_3}COOH}} = \frac{{12}}{{20\% }} = 60{\text{ gam}}\)
