`n_{Mg}=\frac{2,4}{24}=0,1(mol)`
`2CH_3COOH+Mg->(CH_3COO)_2Mg+H_2`
Theo phương trình
`n_{CH_3COOH}=2n_{Mg}=0,2(mol)`
`=>m_{CH_3COOH}=0,2.60=12(g)`
`=>m_{dd CH_3COOH}=\frac{12}{10%}=120(g)`
`a,`
Theo phương trình
`n_{H_2}=n_{Mg}=0,1(mol)`
`=>m_{H_2}=0,1.2=0,2(mol)`
`c,`
`m_{dd}=2,4+120-0,2=122,2(g)`
Theo phương trình
`n_{(CH_3COO)_2Mg}=n_{Mg}=0,1(mol)`
`=>m_{(CH_3COO)_2Mg}=0,1.142=14,2(g)`
`C%_{(CH_3COO)_2Mg}=\frac{14,2}{122,2}.100=11,62%`