`V_{CH_3COOH}=250ml=0,25(l)`
`a)` Phương trình:
`2CH_3COOH+Mg\to(CH_3COO)_2Mg+H_2`
`n_{\text{muối}}=n_{(CH_3COO)_2Mg}=\frac{7,1}{142}=0,05(mol)`
Ta nhận thấy: `n_{Mg}=n_{H_2}=n_{\text{muối}}=0,05(mol)`
`=> m_{Mg}=24.0,05=1,2g`
`=> V_{H_2}=0,05.22,4=1,12(l)`
`b)` Ta có: `n_{CH_3COOH}=2n_{Mg}=0,1(mol)`
`=> CM_{CH_3COOH}=\frac{n}{V}=\frac{0,1}{0,25}=0,4M`