K/lượng axit axetic trong dd:
\(m_{CH_3COOH}=360.5\%=18\left(g\right)\)
\(\Rightarrow n_{CH_3COOH}=\frac{18}{60}=0,3\left(mol\right)\)
\(n_{Mg}=\frac{4,8}{24}=0,2\left(mol\right)\)
Mg + CH3COOH----> H2+ Mg(CH3COO)2
.....1............1....................................1......................(mol)
Vì 0,3>0,2 nên CH3COOH dư.
=> Tính theo Mg.
Mg + CH3COOH----> H2+ Mg(CH3COO)2
...0,2............0,2..............0,2......................0,2......................(mol)
Theo ĐLBTKL: \(m_{Mg}+m_{CH_3COOH}=m_{H_2}+m_{Mg\left(CH_3COO\right)_2}\)
\(m_{dd}=360+4,8-0,2.2=364,4\left(g\right)\)
\(C\%_{ddsauphanung}=\frac{0,2.142}{364,4}.100\approx7,8\%\)
