Gọi số mol Fe, Zn là a, b (mol)

=> 56a + 65b = 12,1 (1)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

             a----------------------->a

            Zn + 2HCl --> ZnCl₂ + H₂

             b----------------------->b

=> a + b =0,2 (2)

(1)(2) => a = 0,1 (mol); b = 0,1 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{12,1}.100\%=46,28\%\\\%m_{Zn}=\dfrac{0,1.65}{12,1}.100\%=53,72\%\end{matrix}\right.\)

- Chất rắn không tan là Cu.

Gọi: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\) ⇒ 27x + 56y = 17,4 - 6,4 (1)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{3}{2}x+y=\dfrac{9,916}{24,79}=0,4\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\end{matrix}\right.\)

Xem lại đề bại ơi

nH2 = 4,48/22,4 = 0,2 (mol)

PTHH: Fe + 2HCl -> FeCl2 + H2

nFe = nH2 = 0,2 (mol)

mFe = 0,2 . 56 = 11,2 (g)

nH2 = 4.48/22.4 = 0.2 (mol) 

Đặt : 

nFe = a (mol) 

nZn = b (mol) 

=> 56a + 65b = 12.1 (1) 

Fe + 2HCl => FeCl2 + H2 

Zn + 2HCl => ZnCl2 + H2 

nH2 = a + b = 0.2 (2) 

(1) , (2) : 

a = b = 0.1 

mFe = 0.1*56 = 5.6 (g) 

mZn = 6.5 (g) 

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

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

\(\Rightarrow56x+65y=12,1\left(1\right)\)

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

Theo PT: \(n_{H_2}=n_{Fe}+n_{Zn}=x+y\left(mol\right)\)

⇒ x + y = 0,2 (2)

Từ (1) và (2) ⇒ x = y = 0,1 (mol)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Zn}=0,1.65=6,5\left(g\right)\end{matrix}\right.\)

Bạn tham khảo nhé!

Bài 6.

\(V=15,68dm^3=15,68l\Rightarrow n_{H_2}=\dfrac{15,68}{22,4}=0,7mol\)

Chất rắn thu đc là \(Cu\) có khối lượng là \(m_{Cu}=16g\)

\(\left\{{}\begin{matrix}Zn:x\left(mol\right)\\Fe:y\left(mol\right)\end{matrix}\right.\Rightarrow65x+56y=56,5-16\left(1\right)\)

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

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(\Rightarrow x+y=n_{H_2}=0,7\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{13}{90}\\y=\dfrac{5}{9}\end{matrix}\right.\)

b)\(\%m_{Cu}=\dfrac{16}{56,5}\cdot100\%=28,31\%\)

\(\%m_{Zn}=\dfrac{\dfrac{13}{90}\cdot65}{56,5}\cdot100\%=16,62\%\)

\(\%m_{Fe}=100\%-\left(28,31\%+16,62\%\right)=55,07\%\)

c)\(n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2mol\)

\(Fe_3O_4+4H_2\rightarrow3Fe+4H_2O\)

0,2            0,7         0           0

0,175        0,7         0,525    0,7

0,025        0            0,525    0,7

\(m_{Fe}=0,525\cdot56=29,4g\)

Bài 1:

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, Giả sử: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)

⇒ 65x + 27y = 21,1 (1)

Ta có: \(n_{H_2}=0,65\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y\left(mol\right)\)

\(\Rightarrow x+\dfrac{3}{2}y=0,65\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,2.65}{21,1}.100\%\approx61,6\text{% }\\\%m_{Al}\approx38,4\%\end{matrix}\right.\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,2.136=27,2\left(g\right)\\m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\end{matrix}\right.\)

Bạn tham khảo nhé!

Bài 2: Cu không tác dụng với dd HCl nên cr thu được là Cu.

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

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

\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)

\(\Rightarrow m_{cr}=m_{Cu}=10-5,6=4,4\left(g\right)\)

Bạn tham khảo nhé!

Bài 3:

PT: \(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

\(2Al+3Cl_2\underrightarrow{t^o}2AlCl_3\)

a, Giả sử: \(\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)

⇒ 56x + 27y = 2,2 (1)

Ta có: \(n_{Cl_2}=0,09\left(mol\right)\)

Theo PT: \(n_{Cl_2}=\dfrac{3}{2}n_{Fe}+\dfrac{3}{2}n_{Al}=\dfrac{3}{2}x+\dfrac{3}{2}y\left(mol\right)\)

\(\Rightarrow\dfrac{3}{2}x+\dfrac{3}{2}y=0,09\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,04\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,02.56}{2,2}.100\%\approx50,91\%\\\%m_{Al}\approx49,09\%\end{matrix}\right.\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_3}=n_{Fe}=0,02\left(mol\right)\\n_{AlCl_3}=n_{Al}=0,04\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{FeCl_3}=0,02.162,5=3,25\left(g\right)\\m_{AlCl_3}=0,04.133,5=5,34\left(g\right)\end{matrix}\right.\)

Bạn tham khảo nhé!

Câu 1:

Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

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

\(n_{Fe}=\dfrac{0,3}{2}=0,15mol\\ m_{Fe}=0,15.56=8,4g\)

Câu 2:

Zn + H₂SO₄ \(\rightarrow\) ZnSO₄ + H₂

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\\ n_{ZnSO_4}=0,6mol\)

\(m_{ZnSO_4}=0,6.161=96,6g\)

Câu 4:

3Fe + 2O₂ \(\underrightarrow{t^o}\) Fe₃O₄

2Cu + O₂ \(\underrightarrow{t^o}\) 2CuO

CaO + H₂O \(\rightarrow\) Ca(OH)₂ 

2Na + 2H₂O \(\rightarrow\) 2NaOH + H₂

NaOH + HCl \(\rightarrow\) H₂O + NaCl

2KOH + H₂SO₄ \(\rightarrow\) K₂SO₄ + 2H₂O

Đáp án B