pt 2CH3COOH+Mg\(\rightarrow\)(CH3COO)2Mg +H2
n(CH3COO)2Mg =1,42/142=0,1 mol
theo pt nCH3COOH =2 n (CH3COO)2Mg =0,1 mol
suy ra Cm=0,2 /0,5=0.4 mol/l
theo pt nH2 =n (CH3COO)2Mg =0,1 mol
suy ra v h2 =2,24l
\(n_{\left(CH_3COO\right)_2Mg}=\dfrac{1,42}{142}=0,01\left(mol\right)\)
Pt: \(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
0,02 <---------------------------0,01 ---------------> 0,01
a) \(C_{M_{axit}}=\dfrac{0,02}{0,05}=0,4M\)
b) \(V_{H_2}=0,01.22,4=0,224l\)