m_HCl = \(\dfrac{146.20\%}{100\%}\)= 29,2 (g)

=> n_HCl = \(\dfrac{29,2}{36,5}\)= 0,8 (mol)

Gọi x,y lần lượt là số mol của Fe₂O₃, MgO

Fe₂O₃ + 6HCl ----> 2FeCl₃ + 3H₂O

x 6x 2x 3x (mol)

MgO + 2HCl ----> MgCl₂ + H₂O

y 2y y y (mol)

Theo PT, ta có:

6x + 2y = 0,8

160x + 40y = 18

=> x = 0,05

y = 0,25

=> m_Fe2O3 = 0,05.160 = 8 (g)

=> %Fe₂O₃ = \(\dfrac{8.100\%}{18}\)= 44,4%

=> % MgO = 100 - 44,4 = 55,6%

b,

m_{dd sau phản ứng} = 18 + 146 = 164 (g)

m_FeCl3 = 0,1.162,5 = 16,25 (g)

m_MgCl2 = 0,25.95 = 23,75 (g)

=> %FeCl₃ = \(\dfrac{16,25.100\%}{164}\)= 10%

=> %MgCl₂ = \(\dfrac{23,75.100\%}{164}\)= 14,48%

\(a)n_{H_2SO_4}=0,15.2=0,3mol\\ FeO+H_2SO_4\rightarrow FeSO_4+H_2O\\ n_{FeO}=n_{FeSO_4}=n_{H_2SO_4}=0,3mol\\ \%m_{FeO}=\dfrac{0,3.72}{22,4}\cdot100\%=96,43\%\\ \%m_{CuO}=100\%-96,43\%=3,57\%\\ b)C_{M_{FeSO_4}}=\dfrac{0,3}{0,15}=2M\)

CuO+2HCl--->CuCl2+H2O

x        2x           x         x

Fe2O3+6HCl--->2FeCl3+3H2O

y            6y          2y          3y

nHCl=3,5.0,2=0,7 mol

gọi x y lần lượt là nCuO,nFe2O3

ta có x.80+y.160=20

         2x+6y=0,7 giải hệ pt ta có  x=0,05   y=0,1

mCuO=0,05.80=4g

mFe2O3=0,1.160=16g

%mCuO=4/20   .100%=20%

%mFe2O3=100%-20%=80%

\(n_{HCl}=0,2.3,5=0,7\left(mol\right)\\ n_{CuO}=a;n_{Fe_2O_3}=b\\ CuO+2HCl\xrightarrow[]{}\Rightarrow CuCl_2+H_2O\\ Fe_2O_3+6HCl\xrightarrow[]{}2FeCl_3+H_2O\\ \left\{{}\begin{matrix}2a+6b=0,7\\80a+160b=20\end{matrix}\right.\\\Rightarrow a=0,05;b=0,1\\ \%_{CuO}=\dfrac{0,05.80}{20}\cdot100=20\%\\ \%_{Fe_2O_3}=\dfrac{0,1.160}{20}\cdot100=80\%\)

a) Pt : \(ZnO+2HCl\rightarrow ZnCl_2+H_2O|\)

             1           2               1           1

             a         2a

              \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O|\)

                   1           6               2             3

                   b          6b

b) Gọi a là số mol của ZnO

           b là số mol của Al₂O₃     
\(m_{ZnO}+m_{Al2O3}=34,5\left(g\right)\)

⇒ \(n_{ZnO}.M_{ZnO}+n_{Al2O3}.M_{Al2O3}=34,5g\)

 ⇒ 81a + 102b = 34,5g

Ta có : 400ml = 0,4l

\(n_{HCl}=3.0,4=1,2\left(mol\right)\)

 ⇒ 2a + 6b = 1,2(2)

Từ (1),(2), ta có hệ phương trình : 

         81a + 102b = 34,5g

            2a + 6b = 1,2

          ⇒ \(\left\{{}\begin{matrix}a=0,3\\b=0,1\end{matrix}\right.\)

\(m_{ZnO}=0,3.81=24,3\left(g\right)\)

\(m_{Al2O3}=0,1.102=10,2\left(g\right)\)

⁰/₀ZnO = \(\dfrac{24,3.100}{34,5}=70,43\)⁰/₀

⁰/₀Al₂O₃ = \(\dfrac{10,2.100}{34,5}=29,57\)⁰/₀

 Chúc bạn học tốt

a/ \(n_{HCl}=0,4.3=1,2\left(mol\right)\)

PTHH: ZnO + 2HCl → ZnCl₂ + H₂O

Mol:       x         2x

PTHH: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O

Mol:        y           6y

b, Ta có: \(\left\{{}\begin{matrix}81x+102y=34,5\\2x+6y=1,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}81\left(0,6-3y\right)+102y=34,5\\x=0,6-3y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,3\\y=0,1\end{matrix}\right.\)

\(\%m_{ZnO}=\dfrac{0,3.81.100\%}{34,5}=70,43\%;\%m_{Al_2O_3}=100-70,43=29,57\%\)

PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

                  a_____2a_______a_______a     (mol)

           \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

                  b_____6b_______2b_______3b   (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}40a+102b=18,2\\2a+6b=\dfrac{182,5\cdot20\%}{36,5}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgO}=\dfrac{0,2\cdot40}{18,2}\cdot100\%\approx43,96\%\\\%m_{Al_2O_3}=56,04\%\end{matrix}\right.\)

Theo PTHH: \(n_{MgCl_2}=0,2\left(mol\right)=n_{AlCl_3}\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,2\cdot95=19\left(g\right)\\m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{19}{18,2+182,5}\cdot100\%\approx9,47\%\\C\%_{AlCl_3}=\dfrac{26,7}{182,5+18,2}\cdot100\%\approx13,3\%\end{matrix}\right.\)

a) mHCl=182,5. 20%=36,5(g) -> nHCl=1(mol)

Đặt nMgO=a(mol); nAl2O3=b(mol)

PTHH: MgO +2 HCl -> MgCl2 + H2O

a__________2a______a(mol)

Al2O3 + 6 HCl ->  2 AlCl3 + 3 H2O

b_______6b______2b(mol)

b) Ta có hpt:

\(\left\{{}\begin{matrix}40a+102b=18,2\\2a+6b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

=> mMgO=0,2.40=8(g)

=>%mMgO=(8/18,2).100=43,956%

=> %mAl2O3= 56,044%

c) m(muối)= mAlCl3 + mMgCl2= 133,5.2b+ 95.a= 133,5.0,1.2+95.0,2= 45,7(g)

d) mAlCl3= 26,7(g) ; mMgCl2 = 19(g)

mddsau= 18,2+ 182,5= 200,7(g)

=>C%ddAlCl3=(26,7/200,7).100=13,303%

C%ddMgCl2=(19/200,7).100=9,467%

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{H_2}=0,2(mol)\\ a,C_{M_{HCl}}=\dfrac{0,2}{0,1}=2M\\ b,n_{Fe}=n_{H_2}=0,1(mol)\\ \Rightarrow m_{Fe}=0,1.56=5,6(g)\\ \Rightarrow m_{Cu}=20-5,6=14,4(g)\\ c,\%m_{Fe}=\dfrac{5,6}{20}.100\%=28\%\\ \%m_{Cu}=100\%-28\%=72\%\)