\(n_{\left(CH_3COO\right)_2Mg}=\dfrac{0,71}{142}=0,005\left(mol\right)\)
PTHH: 2CH3COOH + Mg ---> (CH3COO)2Mg + H2
0,01<----------------------0,005---------->0,005
\(C_{M\left(CH_3COOH\right)}=\dfrac{0,01}{0,025}=4M\)
PTHH: CH3COOH + NaOH ---> CH3COONa + H2O
0,005--------->0,005
\(V_{ddNaOH}=\dfrac{0,005}{0,75}=\dfrac{1}{150}M\\ V_{H_2}=0,005.22,4=0,0112\left(l\right)\)
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