Đáp án:
0,15 l
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2C{H_3}COOH + Mg \to {(C{H_3}COO)_2}Mg + {H_2}\\
2C{H_3}COOH + Fe \to {(C{H_3}COO)_2}Fe + {H_2}\\
n{H_2} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
hh:Mg(a\,mol),Fe(b\,mol)\\
a + b = 0,15\\
24a + 56b = 5,2\\
\Rightarrow a = 0,1;b = 0,05\\
\% mMg = \dfrac{{2,4}}{{5,2}} \times 100\% = 46,15\% \\
\% mFe = 100 - 46,15 = 53,85\% \\
b)\\
nC{H_3}COOH = 2n{H_2} = 0,3\,mol\\
VC{H_3}COOH = \dfrac{n}{{{C_M}}} = \dfrac{{0,3}}{2} = 0,15l
\end{array}\)