Đáp án:
\(\begin{array}{l}
a)\\
{C_\% }C{H_3}COOH = 6\% \\
b)\\
{V_{{H_2}}} = 2,24l\\
c)\\
{C_\% }{(C{H_3}COO)_2}Mg = 7,02\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + 2C{H_3}COOH \to {(C{H_3}COO)_2}Mg + {H_2}\\
{n_{Mg}} = \dfrac{{2,4}}{{24}} = 0,1\,mol\\
{n_{C{H_3}COOH}} = 2{n_{Mg}} = 0,2\,mol\\
{C_\% }C{H_3}COOH = \dfrac{{0,2 \times 60}}{{200}} \times 100\% = 6\% \\
b)\\
{n_{{H_2}}} = {n_{Mg}} = 0,1\,mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
c)\\
{m_{{\rm{dd}}spu}} = 2,4 + 200 - 0,1 \times 2 = 202,2g\\
{n_{{{(C{H_3}COO)}_2}Mg}} = {n_{Mg}} = 0,1\,mol\\
{C_\% }{(C{H_3}COO)_2}Mg = \dfrac{{0,1 \times 142}}{{202,2}} \times 100\% = 7,02\%
\end{array}\)
