\(n_{CO2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

a) Pt : \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O|\)

            1              2             1            1          1

          0,1            0,2          0,1         0,1

           \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

             1           2              1               1

           0,05                      0,05

b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

⇒ \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(m_{dd}=\dfrac{7,3.100}{7,3}=100\left(g\right)\)

d) \(n_{MgCO3}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{MgCO3}=0,1.84=8,4\left(g\right)\)

\(m_{MgO}=10,4-8,4=2\left(g\right)\)

\(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)

\(n_{MgCl2}=0,1+0,05=0,15\left(mol\right)\)

⇒ \(m_{MgCl2}=0,15.95=14,25\left(g\right)\)

Sau phản ứng :

\(m_{dd}=10,4+100-\left(0,1.44\right)\)

       = 106 (g)

\(C_{MgCl2}=\dfrac{14,25.100}{106}=13,44\)⁰/₀

 Chúc bạn học tốt

E xem lại đề 

\(\text{nCO2= 4,48:22,4=0,2 mol}\)

MgO + 2HCl -> MgCl2+ H2O

MgCO3 +2 HCl -> MgCl2+ H2O + CO2

mol 0,2....<- 0,4.....<-0,2 .................<-0,2

\(\Rightarrow\text{ mMgCO3=0,2.84=16,8(g) }\)

\(\Rightarrow\text{mMgO=20,8-16,8=4g}\)

\(\Rightarrow\text{%MgCO3= 16,8:20,8.100%=80,77%}\)

\(\Rightarrow\text{ %MgO=100-80,77=19,23%}\)

\(\text{b, nMgO=4÷ 0,1(mol)}\)

Theo PTHH 1: n HCl= 2n MgO= 0,1.2= 0,2

\(\text{nHCl = 2nMgCO3= 0,2.2= 0,4}\)

\(\Rightarrow\text{nHCl=0,2+0,4=0,6}\)

\(\Rightarrow\text{mHCl=0,6. 36,5=21,9g}\)

\(\Rightarrow\text{mddHCl= 21,9.100:7,3=300g}\)

\(\Rightarrow\text{ V= 300: 1,05=285,71g}\)

MgO+2HCl---->MgCl2+H2

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

n\(_{CO2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

Theo pthh2

n\(_{MgCO3}=n_{CO2}=0,2\left(mol\right)\)

%m MgCO3=\(\frac{0,2.84}{20,8}.100\%=80,77\%\)

%m MgO=100-80,77=19,23%

b) mol HCl ở PT1 là n\(_{HCl}=2n_{MgO}=\frac{4}{40}.2=0,2\left(mol\right)\)

mol HCl ở pt2 là n\(_{HCl}=2n_{CO2}=0,4\left(mol\right)\)

Tổng mol HCl =0,2+0,4=0,6(mol)

m\(_{ddHCl}=\frac{0,6.36,5.100}{7,3}=300\left(g\right)\)

V HCl=300/1,05=285,71(l)

c) Theo pthh1

n\(_{MgCl2}=n_{MgO}=0,1\left(mol\right)\)

m dd sau pư=300+20,8-8,8=312(g)

Theo pthh2

n\(_{MgCl2}=n_{CO2}=0,2\left(mol\right)\)

C% MgCl2 =\(\frac{0,3.95}{312}.100\%=9,13\%\)

\(Zn+2HCl\rightarrow Zn+H_2\)

\(ZnO+2HCl\rightarrow Zn+H_2O\)

Ta có : \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)

\(m_{Zn}=0,1.65=6,5\left(g\right)\)

=> \(\%m_{Zn}=\dfrac{6,5}{14.6}.100=44,52\%\)

=> % m ZnO = 55,48%

Ta có: \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(n_{HCl}=0,4.1,5=0,6\left(mol\right)\)

Giả sử: \(\left\{{}\begin{matrix}n_{Na_2CO_3}=x\left(mol\right)\\n_{K_2CO_3}=y\left(mol\right)\end{matrix}\right.\)

PT: \(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)

\(K_2CO_3+2HCl\rightarrow2KCl+H_2O+CO_2\)

\(HCl_{dư}+NaOH\rightarrow NaCl+H_2O\)

Theo PT: \(n_{CO_2}=n_{Na_2CO_3}+n_{K_2CO_3}=x+y\left(mol\right)\) ⇒ x + y = 0,25 (1)

\(n_{HCl\left(pư\right)}=2x+2y\left(mol\right)\) \(\Rightarrow n_{HCl\left(dư\right)}=0,6-2x-2y\left(mol\right)\)

Có: \(\left\{{}\begin{matrix}n_{NaCl}=2n_{Na_2CO_3}+n_{HCl\left(dư\right)}=0,6-2y\left(mol\right)\\n_{KCl}=2n_{K_2CO_3}=2y\left(mol\right)\end{matrix}\right.\)

⇒ 58,5(0,6 - 2y) + 74,5.2y = 39,9 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na_2CO_3}=\dfrac{0,1.106}{0,1.106+0,15.138}.100\%\approx33,9\%\\\%m_{K_2CO_3}\approx66,1\%\end{matrix}\right.\)

Bạn tham khảo nhé!