Đáp án A

.100 => m_HCl = 43,78 (g)

n_HCl = 1,2 (mol)

Gọi n_Zn = a, n_ZnO = b

Zn + 2HCl → ZnCl₂ + H₂

0,4       0,8   ←            0,4 (mol)

ZnO + 2HCl → ZnCl₂ + H₂O

0,2  ←   0,4                               (mol)

.100%

.100% = 61,61%

%m_ZnO = 100% -61,6% = 38,4%

nH2 = 6,72/22,4 = 0,3 (mol)

PTHH: 2Al + 6HCl -> 2AlCl3 + 3H2

nAl = 0,3 : 3 . 2 = 0,2 (mol)

nHCl (Al) = 0,3 . 2 = 0,6 (mol)

mAl = 0,2 . 27 = 5,4 (g)

%mAl = 5,4/25,65 = 20,05%

%mZnO = 100% - 20,05% = 79,95%

mZnO = 25,65 - 5,4 = 20,25 (g)

nZnO = 20,25/81 = 0,25 (mol)

PTHH: ZnO + 2HCl -> ZnCl2 + H2O

nHCl (ZnO) = 0,25 . 2 = 0,5 (mol)

nHCl (đã dùng) = 0,6 + 0,5 = 1,1 (mol)

CMddHCl = 1,1/0,1008 = 10,9M

C% = (10,9 . 36,5)/(10 . 1,19) = 33,43%

PTHH:

CuO + 2HCl ---> CuCl₂ + H₂O (1)

ZnO + 2HCl ---> ZnCl₂ + H₂O (2)

Gọi x, y lần lượt số mol của CuO và ZnO

Đổi 100ml = 0,1 lít

Ta có: \(C_{M_{HCl}}=\dfrac{n_{HCl}}{0,1}=3M\)

=> n_HCl = 0,3(mol) 

Theo PT₍₁₎: n_HCl = 2.n_CuO = 2x

Theo PT₍₂₎: n_HCl = 2.n_ZnO = 2y

=> 2x + 2y = 0,3 (*)

Theo đề, ta có: 80x + 81y = 12,1 (**)

Từ (*) và (**), ta có HPT:

\(\left\{{}\begin{matrix}2x+2y=0,3\\80x+81y=12,1\end{matrix}\right.\)

=> x = 0,05, y = 0,1

=> m_CuO = 0,05 . 80 = 4(g)

m_ZnO = 0,1 . 81 = 8,1(g)

=> \(\%_{m_{CuO}}=\dfrac{4}{12,1}.100\%=33,06\%\)

\(\%_{m_{ZnO}}=100\%-33,06\%=66,94\%\)

Giúp em vs ạ😢

\(n_{HCl}=0,1.3=0,3\left(mol\right)\)

PTHH: CuO + 2HCl → CuCl₂ + H₂O

Mol:       x           2x

PTHH: ZnO + 2HCl → ZnCl₂ + H₂O

Mol:       y           2y

Ta có: \(\left\{{}\begin{matrix}80x+81y=12,1\\2x+2y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}80\left(0,15-y\right)+81y=12,1\\x+y=0,15\end{matrix}\right.\)

   \(\Leftrightarrow\left\{{}\begin{matrix}12-80y+81y=12,1\\x+y=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0,1\\x=0,05\end{matrix}\right.\)

\(\%m_{CuO}=\dfrac{0,05.80.100\%}{12,1}=33,06\%\)

\(\%m_{ZnO}=100-33,06=66,94\%\)

Zn+2HCl--->ZnCl2+H2

Zn+2HCl--->ZnCl2+H2

Ta có

m dd HCl=100,8.1,19=119,952(g)

m HCl=119,952.36/100=43,18(g)

n HCl=43,18/36,5=1,183(mol)

Mà n H2=8,96/22,4=0,4(mol)

Theo pthh1

n HCl=2n H2=0,8(mol)

m HCl ở Pt 2=1,183-0,8=0,383(mol)

Theo pthh2

n ZnO=1/2n HCl=0,1915(mol)

m ZnO=0,1915.81=15,5115(g)

m Zn=0,4.65=26(g)

m Zn+m ZnO=26+15,5115=41,5115(g)

%m ZnO=15,5115/41,5115.100%=37,37%

Chúc bạn học tốt

PTHH ( I ) : \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

PTHH ( II ) : \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

\(m_{ddHCl}=D_{HCl}.V_{HCl}=100,8.1,19=119,952\left(g\right)\)

=> \(m_{HCl}=\frac{C\%_{HCl}.m_{ddHCl}}{100\%}=\frac{36\%.119,952}{100\%}=43,18272\left(g\right)\)

=> \(n_{HCl}=\frac{m_{HCl}}{M_{HCl}}=\frac{43,18272}{1+35,5}\approx1,18\left(mol\right)\)

Mà \(n_{HCl}=n_{HCl\left(I\right)}+n_{HCl\left(II\right)}\)

=> \(2n_{Zn}+2n_{ZnO}=1,18\) ( I )

\(n_{H_2}=\frac{V_{H_2}}{22,4}=\frac{8,96}{22,4}=0,4\left(mol\right)\)

Theo PTHH ( I ) : \(n_{Zn}=n_{H2}=0,4\left(mol\right)\)

=> \(m_{Zn}=n.M=0,4.65=26\left(g\right)\)

Thay \(n_{Zn}=0,4\) vào phương trình ( I ) ta được :

\(2.0,4+2n_{ZnO}=1,18\)

=> \(n_{ZnO}=0,19\left(mol\right)\)

=> \(m_{ZnO}=n.M=0,19.\left(65+16\right)=15,39\left(g\right)\)

Ta có : \(m_{hh}=m_{Zn}+m_{ZnO}=26+15,39=41,39\left(g\right)\)

=> \(\%ZnO=\frac{15,39}{41,39}.100\%\approx37,18\%\)

\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.2.........0.4.........0.2......0.2\)

\(m_{Zn}=0.2\cdot65=13\left(g\right)\Rightarrow m_{ZnO}=14.6-13=1.6\left(g\right)\)

\(\%Zn=\dfrac{13}{14.6}\cdot100\%=89.04\%\)

\(\%ZnO=100\%-89.04\%=10.96\%\)

\(n_{ZnO}=\dfrac{1.6}{81}\approx0.02\left(mol\right)\)

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

\(0.02........0.04........0.02........0.02\)

\(n_{HCl}=0.4+0.04=0.44\left(mol\right)\)

\(C_{M_{HCl}}=\dfrac{0.44}{0.8}=0.55\left(M\right)\)

Eeeee ngồi tính sang chấn thật nó ra số xấu lần mò hơn 20p chưa biết tính sai chỗ nào

m_{dd HCl} = 100,8.1,19 = 119,952 (g)

=> m_HCl = \(\dfrac{119,952.36,5\%}{100\%}\)= 43,8 (g)

=> n_HCl = \(\dfrac{43,8}{36,5}\)= 1,2 (mol)

Zn + 2HCl ----> ZnCl₂ + H₂

0,4 0,8 0,4 0,4 (mol)

=> m_Zn = 0,4.65 = 26 (g)

ZnO + HCl ----> ZnCl₂ + H₂O

0,4 0,4 0,4 0,4 (mol)

=> m_ZnO = 0,4.81 = 32,4 (g)

=> m_{hỗn hợp ban đầu} = 26 + 32,4 = 58,4 (g)

=> %Zn = \(\dfrac{26.100\%}{58,4}\)= 44,52%

=> %ZnO = \(\dfrac{32,4.100\%}{58,4}\)= 55,48%

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

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

a) Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)=n_{Zn}\)

\(\Rightarrow\%m_{Zn}=\dfrac{0,4\cdot65}{36,2}\cdot100\%\approx71,23\%\) \(\Rightarrow\%m_{Al_2O_3}=28,77\%\)

c) Ta có: \(n_{Al_2O_3}=\dfrac{36,2-0,4\cdot65}{102}=0,1\left(mol\right)\)

Theo PTHH: \(n_{HCl}=2n_{Zn}+6n_{Al_2O_3}=1,4\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{1,4\cdot36,5}{10\%}=511\left(g\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{511}{1,1}\approx464,5\left(ml\right)=0,4645\left(l\right)\)

c) Theo PTHH: \(\left\{{}\begin{matrix}n_{ZnCl_2}=0,4\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnCl_2}}=\dfrac{0,4}{0,4645}\approx0,86\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4645}\approx0,43\left(M\right)\end{matrix}\right.\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

Ta có:

\(n_{H2}=n_{Zn}=\frac{8,96}{22,4}=0,4\left(mol\right)\Rightarrow m_{Zn}=0,4.65=26\left(g\right)\)

m dung dịch HCl=100,9.1,19=120 gam

-> m HCl=120.36,5%=43,8 gam

-> nHCl=43,8/36,5=1,2 mol

nHCl=2nZn + 2nZnO

-> nZnO=0,2 mol -> mZnO=0,2.(65+16)=16,2 gam

nZnCl2=nZnO + nZn=0,6 mol

-> mZnCl2=0,6.(65+35,5.2)=81,6 gam

BTKL: m rắn + m dung dịch HCl= m dung dịch sau phản ứng + mH2

-> m dung dịch sau phản ứng = 26+16,2+120-0,4.2=161,4 gam

-> C% ZnCl2=81,6/161,4=50,55%

.100 => mHCl = 43,78 (g)

nHCl = 1,2 (mol)

Gọi nZn = a, nZnO = b

Zn + 2HCl → ZnCl2 + H2

0,4 0,8 ← 0,4 (mol)

ZnO + 2HCl → ZnCl2 + H2O

0,2 ← 0,4 (mol)

=>mZn=0,4.65=26 g

%mZnO = 0,2.81=16,2g

mZnCl2=(0,2+0,4).136=81,6g

a) Đặt \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow24a+27b=5,1\)  (1)

Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Bảo toàn electron: \(2a+3b=0,5\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1\cdot24}{5,1}\cdot100\%\approx47,06\%\\\%m_{Al}=52,94\%\end{matrix}\right.\)

b) Bảo toàn nguyên tố: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5\cdot36,5}{7,3\%}=250\left(g\right)\)

\(\Rightarrow V_{HCl}=\dfrac{250}{1,2}\approx208,33\left(ml\right)\)