a) 

\(n_{MgCl_2}=\dfrac{38}{95}=0,4\left(mol\right)\)

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: MgCO₃ + 2HCl --> MgCl₂ + CO₂ + H₂O

                0,3<------0,6<------0,3<----0,3

            MgO + 2HCl --> MgCl₂ + H₂O

               0,1<---0,2<------0,1

=> \(\left\{{}\begin{matrix}m_{MgO}=0,1.40=4\left(g\right)\\m_{MgCO_3}=0,3.84=25,2\left(g\right)\end{matrix}\right.\)

b) \(m_{HCl}=\left(0,6+0,2\right).36,5=29,2\left(g\right)\)

=> \(m_{dd.HCl}=\dfrac{29,2.100}{20}=146\left(g\right)\)

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!

a/ \(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(MgCO_3+H_2SO_4\rightarrow MgSO_4+H_2O+CO_2\)

\(0,5---0,5----0,5---0,5-0,5\)

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

\(b---b----b-----b\)

\(\Rightarrow m_{MgCO_3}=0,5.\left(24+12+16.3\right)=42\left(g\right)\)

\(\dfrac{m_{MgCO_3}}{m_{MgO}}=\dfrac{7}{3}\Rightarrow m_{MgO}=42.\dfrac{3}{7}=18\left(g\right)\Rightarrow n_{MgO}=b=0,45\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4}=0,45+0,5=0,95\left(mol\right)\) \(\Rightarrow m_{dd}=\dfrac{0,95.98}{0,05}=1862\left(g\right)\)

\(a.PTHH:\)

\(Mg+2HCl--->MgCl_2+H_2\uparrow\left(1\right)\)

\(MgO+2HCl--->MgCl_2+H_2O\left(2\right)\)

b. ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT₍₁₎: \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{MgO}=12-0,2.24=7,2\left(g\right)\)

\(\Rightarrow\%_{MgO}=\dfrac{7,2}{12}.100\%=60\%\)

c. Ta có: \(n_{hh}=0,2+\dfrac{7,2}{40}=0,38\left(mol\right)\)

Theo PT_(1,2): \(n_{HCl}=2.n_{hh}=2.0,38=0,76\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,76.36,5=27,74\left(g\right)\)

\(\Rightarrow m_{dd_{HCl}}=138,7\left(g\right)\)

\(\Rightarrow V_{dd_{HCl}}=126\left(ml\right)\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)

\(Ca+2HCl\rightarrow CaCl_2+H_2\)

0,1                                  0,1     ( mol )

\(\rightarrow\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,1.40}{10}.100=40\%\\\%m_{MgO}=100\%-40\%=60\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{0,1.111}{10+390,2-0,1.2}.100=2,775\%\\C\%_{MgO}=\dfrac{4}{10+390,2-0,1.2}.100=1\%\end{matrix}\right.\)

Chọn C