Question

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Answer 1

\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\)

PTHH: \(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\uparrow\)

                0,5<-----------0,25------->0,25------------->0,25

\(\rightarrow C_{M\left(CH_3COOH\right)}=\dfrac{0,5}{0,05}=10M\)

\(m_{dd}=100+6-0,25.2=105,5\left(g\right)\\ \rightarrow C\%_{\left(CH_3COO\right)_2Mg}=\dfrac{0,25.142}{105,5}.100\%=33,65\%\)