\(n_{Al}=\dfrac{2,7}{27}=0,1(mol)\\ PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow n_{HCl}=3n_{Al}=0,3(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,3}{0,5}=0,6(l)\)

Bảo toàn nguyên tố Cl, Al:

\(n_{HCl}=n_{Cl}=3n_{AlCl_3}=3n_{Al}=0,3\left(mol\right)\)

\(\Rightarrow V=\dfrac{0,3}{0,5}=0,6\left(l\right)\)

`1)`

`n_{Al}={2,7}/{27}=0,1(mol)`

`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`

`0,1->0,15->0,05->0,15(mol)`

`V_{dd\ H_2SO_4}={0,15}/1=0,15(l)=150(ml)`

`->V=150`

`V'=V_{H_2}=0,15.22,4=3,36(l)`

`C_{M\ X}=C_{M\ Al_2(SO_4)_3}={0,05}/{0,15}=1/3M`

`2)`

`n_{Fe}={2,8}/{56}=0,05(mol)`

`Fe+2HCl->FeCl_2+H_2`

`0,05->0,1->0,05->0,05(mol)`

`V_{dd\ HCl}={0,1}/1=0,1(l)=100(ml)`

`->V=100`

`V_{H_2}=0,05.22,4=1,12(l)`

`C_{M\ FeCl_2}={0,05}/{0,1}=0,5M`

\(n_{Fe}=\dfrac{5,6}{56}=0,1(mol);n_{HCl}=1.0,3=0,3(mol)\\ Fe+2HCl\to FeCl_2+H_2\\ \)

Vì \(\dfrac{n_{Fe}}{1}<\dfrac{n_{HCl}}{2}\) nên HCl dư

Do đó \(n_{H_2}=n_{Fe}=0,1(mol)\)

\(\Rightarrow V_{H_2}=0,1.22,4=2,24(l)\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right);n_{HCl}=0,1.1=0,1\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\) => HCl hết, Fe dư

PTHH: Fe + 2HCl --> FeCl₂ + H₂

___________0,1-------------->0,05_____(mol)

=> V_H2 = 0,05.22,4 = 1,12(l)

Có lẽ đề cho dd HCl 1M (1 mol/l) chứ bạn nhỉ?

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

b, \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)\)

c, \(n_{HCl}=3n_{Al}=0,3\left(mol\right)\Rightarrow V_{ddHCl}=\dfrac{0,3}{1}=0,3\left(l\right)\)

n_Al = \(\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\) ( mol )

               2Al   +   6HCl   →   2AlCl₃   +   3H₂

( mol )  0,2                       →    0, 2

\(m_{AlCl_3}=n.M=0,2.\left(27+35,5\times3\right)=26,7\left(g\right)\)

a, \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

 \(m_{HCl}=109,5.10\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:      0,1      0,3         0,1       0,15

Ta có: \(\dfrac{0,1}{2}=\dfrac{0,3}{6}\) ⇒ Al hết, HCl hết

\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

b, \(m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)

c, m_{dd sau pứ} = 2,7 + 109,5 - 0,15.2 = 111,9 (g)

\(C\%_{ddAlCl_3}=\dfrac{13,35.100\%}{111,9}=11,93\%\)

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

Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\n_{HCl}=\dfrac{109,5\cdot10\%}{36,5}=0,3\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{2}=\dfrac{0,3}{6}\) \(\Rightarrow\) Al và HCl đều p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{AlCl_3}=0,1\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,1\cdot133,5=13,35\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Al}+m_{ddHCl}-m_{H_2}=111,9\left(g\right)\)

\(\Rightarrow C\%_{AlCl_3}=\dfrac{13,35}{111,9}\cdot100\%\approx11,93\%\)

\(a)2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b)n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\\ n_{Al}=a;n_{Fe}=b\\ \left\{{}\begin{matrix}3a+b=0,25\\27a+56b=8,3\end{matrix}\right.\\ a=\dfrac{19}{470};b=\dfrac{121}{940}\\ \%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{8,3}\cdot100=13,15\%\\ \%m_{Fe}=100-13,15=86,85\%\\ c)n_{HCl}=3\cdot\dfrac{19}{470}+2\cdot\dfrac{121}{940}=\dfrac{89}{235}mol\\ m_{ddHCl=}=\dfrac{\dfrac{89}{235}\cdot36,5}{7,3}\cdot100=189g\\ d)n_{AlCl_3}=n_{Al}=\dfrac{19}{470}mol\\ n_{Fe}=n_{FeCl_2}=\dfrac{121}{940}mol\)

\(m_{dd}=8,3+189-0,25.2=196,8g\\ C_{\%AlCl_3}=\dfrac{\dfrac{19}{470}\cdot133,8}{196,8}\cdot100=2,8\%\\ C_{\%FeCl_2}=\dfrac{\dfrac{121}{940}127}{196,8}\cdot100=8,3\%\)