\(Đặt:n_{Al}=a\left(mol\right),n_{Fe}=b\left(mol\right)\)

\(m_{hh}=27a+56b=8.3\left(g\right)\left(1\right)\)

\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

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

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

\(Tathấy:\)

\(n_{HCl}=2n_{H_2}=2\cdot0.25=0.5\left(mol\right)\)

\(V_{ddHCl}=\dfrac{0.5}{0.2}=2.5\left(l\right)\)

\(n_{H_2}=1.5a+b=0.25\left(mol\right)\left(2\right)\)

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

\(C_{M_{AlCl_3}}=\dfrac{0.1}{2.5}=0.04\left(M\right)\)

\(C_{M_{FeCl_2}}=\dfrac{0.1}{2.5}=0.04\left(M\right)\)

Chúc em học tốt !!!

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

BTNT H, có: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{0,5}{0,2}=2,5\left(l\right)\)

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

⇒ 27x + 56y = 8,3 (1)

Các quá trình:

\(Al^0\rightarrow Al^{+3}+3e\)

x___________ 3x (mol)

\(Fe^0\rightarrow Fe^{+2}+2e\)

y____________2y (mol)

\(2H^++2e\rightarrow H_2^0\)

______0,5__0,25 (mol)

Theo ĐLBT mol e, có: 3x + 2y = 0,5 (2)

Từ (1) và (2) ⇒ x = y = 0,1 (mol)

BTNT Al và Fe, có: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\\n_{FeCl_3}=n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow C_{M_{AlCl_3}}=C_{M_{FeCl_3}}=\dfrac{0,1}{2,5}=0,04M\)

Bạn tham khảo nhé!

hình như đề số xấu lắm bạn

a.

Fe  +  2HCl → FeCl₂  + H₂

Ag không phản ứng với dung dịch HCl

nH₂ =\(\dfrac{2,24}{22,4}\)=0,1 mol => theo tỉ lệ phản ứng nFe = 0,1 mol và nHCl phản ứng = 0,2 mol 

=> C_HCl =\(\dfrac{0,2}{0,2}\) = 1 ( mol/l)

b.

mFe = 0,1.56 = 5,6 gam => mAg = 16,4-5,6 = 10,8 gam

%mFe = \(\dfrac{5,6}{16,4}.100\)= 34,14% => %mAg = 100- 34,14 = 65,86%

\(n_{Fe}=a\left(mol\right),n_{Mg}=b\left(mol\right)\)

\(m_{hh}=56a+24b=10.16\left(g\right)\)

\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

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

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(n_{H_2}=a+b=0.25\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.13,b=0.12\)

\(m_{Fe}=0.13\cdot56=7.28\left(g\right)\)

\(m_{Mg}=0.12\cdot24=2.88\left(g\right)\)

\(n_{HCl}=2\cdot n_{H_2}=2\cdot0.25=0.5\left(mol\right)\)

\(C_{M_{HCl}}=\dfrac{0.5}{0.5}=1\left(M\right)\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

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

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\\%m_{Fe}=\dfrac{0,1\cdot56}{12}\cdot100\%\approx46,67\%\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\\\%m_{Cu}=53,33\%\end{matrix}\right.\)

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

Cu không phản ứng

\(nH_2=nFe=\dfrac{2,24}{22,4}=0,1mol\)

\(\rightarrow mFe=0,1.56=5,6gam\)

\(\rightarrow\%mFe=\dfrac{5,6}{12}.100\%=46,\left(6\right)\%\)

\(\rightarrow\%mCu=100\%-46,\left(6\right)\%=53,\left(3\right)\%\)

c)

\(CM_{HCl}=\dfrac{0,1.2}{0,2}=1M\)

a) $Zn + 2HCl \to ZnCl_2 + H_2$

$ZnO + 2HCl \to ZnCl_2 + H_2O$

b)

Theo PTHH : $n_{Zn} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$

$m_{Zn} = 0,2.65 = 13(gam)$

$m_{ZnO} = 21,1 - 13 = 8,1(gam)$

c) $n_{ZnO} = 0,1(mol)$

Theo PTHH : $n_{HCl} = 2n_{Zn} + 2n_{ZnO} = 0,6(mol)$
$m_{dd\ HCl} = \dfrac{0,6.36,5}{16,6\%} = 132(gam)$

d) $m_{dd\ sau\ pư} = 21,1 + 132 - 0,2.2 = 152,7(gam)$
$n_{ZnCl_2} = n_{Zn} + n_{ZnO} = 0,3(mol)$

$C\%_{ZnCl_2} = \dfrac{0,3.136}{152,7}.100\% = 26,72\%$

0,2.2 ở đâu  ra vậy ạ

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow n_{Fe}=0,05\left(mol\right)\\ \Rightarrow m_{Fe}=0,05\cdot56=2,8\left(g\right)\\ \Rightarrow\%_{Fe}=\dfrac{2,8}{6}\cdot100\%\approx46,67\%\\ \Rightarrow\%_{Cu}\approx100\%-46,67\%=53,33\%\)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\)

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

 0,05                              0,05

\(m_{Fe}=0,05\cdot56=2,8g\)

\(\%m_{Fe}=\dfrac{2,8}{6}\cdot100\%=46,67\%\)

nFe = 5.6/56 = 0.1 (mol) 

nHCl = 0.2*2 = 0.4 (mol) 

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

LTL : 0.1/1 < 0.4/2 => HCl dư 

mHCl dư = ( 0.4 - 0.2 ) * 36.5 = 7.3 (g) 

VH2 = 0.2*22.4 = 4.48 (l) 

CM FeCl2 = 0.1/0.2 = 0.5(M)

CM HCl dư = 0.2 / 0.2 = 1(M) 

C_{M FeCl2} = n/v = 0,1 / 2,24 = 0,446 chứ nhỉ

\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

____0,35_____0,7___________0,35 (mol)

a, \(m_{Zn}=0,35.65=22,75\left(g\right)\)

b, \(C\%_{HCl}=\dfrac{0,7.36,5}{200}.100\%=12,775\%\)

\(a)n_{H_2}=\dfrac{7,84}{22,4}=0,35mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,35        0,7            0,35         0,35

\(m_{Zn}=0,35.65=22,75g\\ b)C_{\%HCl}=\dfrac{0,7.36,5}{200}\cdot100\%=12,775\%\)