Đáp án:
a) \({{\text{m}}_{Mg}} = 0,48{\text{ gam}}\)
b) \(C{\% _{C{H_3}COOH}} = 27,78\% \)
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:
\({m_{{{(C{H_3}COO)}_2}Mg}} = 2,84 \to {n_{Mg}} = {n_{{{(C{H_3}COO)}_2}Mg}} = \frac{{2,84}}{{59.2 + 24}} = 0,02{\text{ mol}} \to {{\text{m}}_{Mg}} = 0,02.24 = 0,48{\text{ gam}}\)
Phản ứng xảy ra:
\(C{H_3}COOH + KOH\xrightarrow{{}}C{H_3}COOK + {H_2}O\)
Ta có: \({n_{C{H_3}COOH}} = 2{n_{Mg}} = 0,04{\text{ mol}} \to {{\text{m}}_{C{H_3}COOH}} = 0,04.60 = 2,4\;{\text{gam;}}{{\text{m}}_{dd\;{\text{C}}{{\text{H}}_3}COOH}} = 8.1,08 = 8,64{\text{gam}}\)
\( \to C{\% _{C{H_3}COOH}} = \frac{{2,4}}{{8,64}} = 27,78\% \)
