$n_{CO2}$=$\frac{2,24}{22,4}$=0,1 (mol)
a,
2CH3COOH + MgO -> H2O + (CH3COO)2Mg
MgCO3 + 2CH3COOH -> (CH3COO)2Mg + CO2 + H2O
0,1 0,1
b,
$m_{MgCO2}$=0,1.84=8,4 (g)
$m_{MgO}$=8,8-8,4=0,4 (g)
c,
%$m_{MgCO2}$=$\frac{8,4.100}{8,8}$≈95,45%
%$m_{MgO}$=100%-95,45%=4,55%
Cho mình ctlhn cho team nhé.