Đáp án:
\({m_{dd{\text{ C}}{{\text{H}}_3}COOH}} = 100{\text{ gam}}\)
\({V_{{H_2}}} = 11,2{\text{ lít}}\)
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:
\({n_{Mg}} = \frac{{12}}{{24}} = 0,5{\text{ mol = }}{{\text{n}}_{{H_2}}}\)
\({n_{C{H_3}COOH}} = 2{n_{Mg}} = 0,5.1 = 1{\text{ mol}}\)
\( \to {m_{C{H_3}COOH}} = 1.60 = 60{\text{ gam}}\)
\( \to {m_{dd{\text{ C}}{{\text{H}}_3}COOH}} = \frac{{60}}{{60\% }} = 100{\text{ gam}}\)
\({V_{{H_2}}} = 0,5.22,4 = 11,2{\text{ lít}}\)