Ta có:
\(n_{Zn}=\frac{13}{65}=0,2\left(mol\right)\)
\(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
\(\Rightarrow n_{CH3COOH}=2n_{Zn}=0,2.2=0,4\left(mol\right)\)
\(m_{dd\left(CH3COOH\right)}=\frac{0,4.60}{12\%}=200\left(g\right)\)
\(n_{H2}=n_{Zn}=0,2\left(mol\right)\Rightarrow m_{dd\left(spu\right)}=212,6\left(g\right)\)
\(n_{\left(CH3COO\right)2Zn}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow C\%_{\left(CH3COO\right)2Zn}=\frac{0,2.183}{212,6}.100\%=17,22\%\)
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