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

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)

\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

\(n_{HCl}=0.35\cdot1=0.35\left(mol\right)\)

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

Lập tỉ lệ : 

\(0.1< \dfrac{0.35}{2}\Rightarrow HCldư\)

\(m_{FeCl_2}=0.1\cdot127=12.7\left(g\right)\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{Fe}=0,2(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,2}{0,5}=0,4(l)\)

Bảo toàn nguyên tố Cl, Fe:

\(n_{HCl}=2n_{FeCl_2}=2n_{Fe}=0,2\left(mol\right)\)

\(\Rightarrow V=0,4\left(l\right)\)

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

Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\) \(\Rightarrow n_{H_2}=0,15\left(mol\right)\) \(\Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)

Giúp mình với mn :)

\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

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

\(0.1.....................0.1\)

\(m_{FeCl_2}=0.1\cdot127=12.7\left(g\right)\)

Bảo toàn nguyên tố Fe:

\(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=12,7\left(g\right)\)

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

               2Al   +   6HCl   →   2AlCl₃   +   3H₂

( mol )  0,2                       →    0, 2

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

pH=12 => n_OH-= 10^-2 * 0,3 = 0,003 (mol)

để trung hòa thì: n_H+ = n_OH- = 0,003 => Vdd HCl = 0,003/10^-3 = 3 (l)

dd sau khi cô cạn: K⁺ a (mol) , Na⁺ b (mol) , Cl^- 0,003 (mol)

lập đc hệ:             a + b = 0,003

                        39a + 23b = 0,1915 - 0,003 * 35,5

=> a= 0,001 , b= 0,002