Đáp án:
a) 3%
b) 1,12l
c) 4 g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2C{H_3}COOH + Mg \to {(C{H_3}COO)_2}Mg + {H_2}\\
n{(C{H_3}COO)_2}Mg = \dfrac{{7,1}}{{142}} = 0,05\,mol\\
\Rightarrow nC{H_3}COOH = 0,05 \times 2 = 0,1\,mol\\
C\% C{H_3}COOH = \dfrac{{0,1 \times 60}}{{200}} \times 100\% = 3\% \\
b)\\
n{H_2} = n{(C{H_3}COO)_2}Mg = 0,05\,mol\\
\Rightarrow V{H_2} = 0,05 \times 22,4 = 1,12l\\
c)\\
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O\\
nNaOH = nC{H_3}COOH = 0,1\,mol\\
\Rightarrow mNaOH = 0,1 \times 40 = 4g
\end{array}\)