Đáp án:
\( {C_{M{\text{ C}}{{\text{H}}_3}COOH}} = \frac{2}{3}M\)
\({V_{{H_2}}} = 1,12{\text{ lít}}\)
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:
\({n_{{{(C{H_3}COO)}_2}Mg}} = \frac{{7,1}}{{59.2 + 24}} = 0,05{\text{ mol = }}{{\text{n}}_{{H_2}}}\)
\({n_{C{H_3}COOH}} = 2{n_{{{(C{H_3}COO)}_2}Mg}} = 0,05.2 = 0,1{\text{ mol}}\)
\( \to {C_{M{\text{ C}}{{\text{H}}_3}COOH}} = \frac{{0,1}}{{0,15}} = \frac{2}{3}M\)
\({V_{{H_2}}} = 0,05.22,4 = 1,12{\text{ lít}}\)
