Mg + 2CH3COOH => (CH3COO)2Mg + H2
n muối = m/M = 1.42/142 = 0.01 (mol)
Theo phương trình ==> nCH3COOH = 0.02 (mol). 50ml = 0.05 (l)
CM dd CH3COOH = n/V = 0.02/0.05 = 0.4 M
CH3COOH + NaOH => CH3COONa + H2O
50ml---> 0.02 mol; 100ml => 0.04 mol
==> nNaOH = 0.04 (mol)
V dd NaOH = n/Cm = 0.04/0.5 = 0.08 l = 80ml
n(CH3COO)2Mg= 0.01 (mol)
2CH3COOH + Mg --> (CH3COO)2Mg + H2
nCH3COOH= 0.02mol
CM CH3COOH= 0.02/0.05=0.4M
c) n CH3COOH= 0.4*0.1=0.04 mol
CH3COOH + NaOH --> CH3COONa + H2O
=> nNaOH= 0.04 (mol)
VNaOH= 0.04/0.5=0.08 l