\(a) Mg + 2CH_3COOH \to (CH_3COO)_2Mg + H_2\\ n_{H_2} = n_{Mg} = \dfrac{9,6}{24} = 0,4(mol)\\ V_{H_2} = 0,4.22,4 = 8,96(lít)\\ b) n_{(CH_3COO)_2Mg} = n_{Mg} = 0,4(mol)\\ m_{Muối} = 0,4.142 = 56,8(gam)\\ c) n_{CH_3COOH} = 2n_{Mg} = 0,8(mol)\\ m_{dd\ CH_3COOH} = \dfrac{0,8.60}{6\%} = 800(gam)\\ d) C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O\\ n_{C_2H_5OH} = n_{CH_3COOH} = 0,8(mol)\\ m_{C_2H_5OH} = 0,8.46 = 36,8(gam)\)
mMg=9,6/24=0,4 mol
Mg + 2CH3COOH --> (CH3COO)2Mg+H2
0,4 0,8 0,4 0,4 mol
=>VH2=0,4*22,4=8,96 lít
=>m(CH3COO)2Mg=0,4*142=56,8 g
mddCH3COOH=0,8*60/6%=800 g
C2H5OH +O2 -men giấm->CH3COOH + H2O
0,8 0,8 mol
mC2H5OH=0,8*46=36,8G
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