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%

a,\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

PTHH: CuO + 2HCl → CuCl₂ + H₂O

Mol:       x          2x

PTHH: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

Mol:        y            6y

Ta có: \(\left\{{}\begin{matrix}80x+160y=16\\2x+6y=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

PTHH: CuO + 2HCl → CuCl₂ + H₂O

Mol:      0,1       0,2

PTHH: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

Mol:      0,05         0,3

\(\Rightarrow m_{CuO}=0,1.80=8\left(g\right);m_{Fe_2O_3}=16-8=8\left(g\right)\)

b,\(\%m_{CuO}=\dfrac{8.100\%}{16}=50\%;\%m_{Fe_2O_3}=100-50=50\%\)

a) 2HCl + CuO ----> CuCl2 + H2O 



6HCl + Fe2O3 ----> 2FeCl3 + 3H2O 
b) nHCl= 3.5 x 0.2 = 0.7 

Đặt x, y lần lượt là số mọl của HCl ở pt 1, pt2 
2HCl + CuO ----> CuCl2 + H2O 
2x-------------x-----------x--------- x 


6HCl + Fe2O3-----> 2FeCl3 + 3H2O 
6y---------------y----------------2y--... 3y 
ta có hệ phương trình hai ẩn x, y 
2x+ 6y = 0.7 
80x+160y=20 
===> x=0.05;y = 0.1 
m CuO= 0.05 x 80=4 g 
m Fe2O3= 0.1 x 160 =16 g 

\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

a)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

x------->2x

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

y--------->6y

Có hệ: \(\left\{{}\begin{matrix}2x+6y=0,5\\80x+160y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

\(m_{CuO}=0,1.80=8\left(g\right)\\ m_{Fe_2O_3}=0,05.160=8\left(g\right)\)

b

\(\%m_{CuO}=\dfrac{0,1.80.100\%}{16}=50\%\\ \%m_{Fe_2O_3}=\dfrac{0,05.160.100\%}{16}=50\%\)

Đặt: nMgO=x(mol) ; nCuO=y(mol) (x,y>0)

nHCl=0,2.2=0,4(mol)

PTHH: MgO +2 HCl -> MgCl2 + H2

x________2x__________x___x(mol)

CuO +2 HCl -> CuCl2 + H2

y___2y______y_____y(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}40x+80y=10\\2x+2y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,05\end{matrix}\right.\)

=> mMgO= 0,15.40=6(g); mCuO= 0,05.80=4(g)

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

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

Câu 3:

Gọi x, y lần lượt là số mol của MgO và Al₂O₃

Ta có: \(n_{H_2SO_4}=0,2.250:1000=0,05\left(mol\right)\)

a. PTHH: 

MgO + H₂SO₄ ---> MgSO₄ + H₂O (1)

Al₂O₃ + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂O (2)

b. Theo PT₍₁₎: \(n_{H_2SO_4}=n_{MgO}=x\left(mol\right)\)

Theo PT₍₂₎: \(n_{H_2SO_4}=3.n_{Al_2O_3}=3y\left(mol\right)\)

=> x + 3y = 0,05 (1)

Theo đề, ta có: 40x + 102y = 1,82 (2)

Từ (1) và (2), ta có HPT:

\(\left\{{}\begin{matrix}x+3y=0,05\\40x+102y=1,82\end{matrix}\right.\)

=> x = 0,02, y = 0,01

=> \(m_{MgO}=0,02.40=0,8\left(mol\right)\)

=> \(\%_{m_{MgO}}=\dfrac{0,8}{1,82}.100\%=43,96\%\)

\(\%_{m_{Al_2O_3}}=100\%-43,96\%=56,04\%\)

Câu 4:

Ta có: \(m_{H_2SO_4}=\dfrac{19,6\%.100\%}{100}=19,6\left(g\right)\)

=> \(n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)

Ta lại có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

a. PTHH: CuO + H₂SO₄ ---> CuSO₄ + H₂O

Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\)

Vậy H₂SO₄ dư.

Theo PT: \(n_{CuSO_4}=n_{CuO}=0,1\left(mol\right)\)

=> \(m_{CuSO_4}=0,1.160=16\left(g\right)\)

Ta có: \(m_{dd_{CuSO_4}}=8+100=108\left(g\right)\)

=> \(C_{\%_{CuSO_4}}=\dfrac{16}{108}.100\%=14,81\%\)

Câu 5: Thiếu đề

\(n_{CuO}=a\left(mol\right),n_{Fe_2O_3}=b\left(mol\right)\)

\(m=80a+160b=20\left(g\right)\left(1\right)\)

\(n_{HCl}=0.2\cdot3.5=0.7\left(mol\right)\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(n_{HCl}=2a+6b=0.7\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.05,b=0.1\)

\(m_{CuO}=0.05\cdot80=4\left(g\right)\)

\(m_{Fe_2O_3}=0.1\cdot160=16\left(g\right)\)