$n_{CO2}$=$\frac{2,24}{22,4}$=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.

Giải thích các bước giải:

\(\begin{array}{l}
2C{H_3}{\rm{COO}}H + MgO \to {(C{H_3}{\rm{COO)}}_2}Mg + {H_2}O\\
2C{H_3}{\rm{COO}}H + MgC{O_3} \to {(C{H_3}{\rm{COO)}}_2}Mg + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = 0,1mol = {n_{MgC{O_3}}}\\
 \to {m_{MgC{O_3}}} = 8,4g \to {m_{MgO}} = 8,8 - 8,4 = 0,4g\\
 \to \% {m_{MgC{O_3}}}\dfrac{{8,4}}{{8,8}} \times 100\%  = 95,45\% \\
 \to \% {m_{MgO}} = 4,55\% 
\end{array}\)