MgO + 2 HCl -> MgCl2 + H2O
x______2x_____x_______x(mol)
MgCO3 + 2 HCl -> MgCl2 + CO2 + H2O
y___2y__________y______y(mol)
nMgCl2= 38/95=0,4(mol)
nCO2= 6,72/22,4= 0,3(mol)
\(\left\{{}\begin{matrix}y=0,3\\x+y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,3\end{matrix}\right.\)
=> mMgO= 40.0,1=4(g)
mMgCO3= 0,3. 84= 25,2(g)
=> \(\%mMgO=\frac{4}{4+25,2}.100\approx13,699\%\\ \rightarrow\%mMgCO3\approx100\%-13,699\%\approx86,301\%\)
\(\left\{{}\begin{matrix}MgO\\MgCO_3\end{matrix}\right.+HCl\rightarrow\left\{{}\begin{matrix}MgCl_2+H_2O\left(1\right)\\MgCl_2+H_2O+CO_2\left(2\right)\end{matrix}\right.\)
Ta có: \(n_{MgCl_2}=0,4\left(mol\right)\)
\(n_{CO_2}=0,3\left(mol\right)\)
\(\Rightarrow n_{MgCO_3}=n_{MgCl_2\left(2\right)}=0,3\left(mol\right)\)
\(\Rightarrow n_{MgO}=n_{MgCl_2\left(1\right)}=0,1\left(mol\right)\)
\(\Rightarrow\%m_{MgO}=\frac{0,1.40}{0,1.40+0,3.84}.100\%=13,99\text{%}\)
\(\%m_{MgCO_3}=86,01\%\)
Bạn than khảo nhé!