a) \(PTHH:Fe+HCL\) → \(FeCl_2+H_2\)

Cân bằng: \(Fe+2HCl\) → \(FeCl_2+H_2\)

b) \(n_{Fe}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(n_{HCl}=2.n_{Fe}=2.0,1=0,2\left(mol\right)\)

\(m_{HCl}=n.M=0,2.36,5=7,3\left(g\right)\)

c) \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)

\(V_{H_2\left(đktc\right)}=n.22,4=0,1.22,4=2,24\left(l\right)\)

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\) 

_________0,1__________0,1

\(b.n_{H_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ \Rightarrow m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

\(\text{Ta có PTHH}\\2Al+6HCl \rightarrow 2AlCl_3+3H_2 \uparrow\\n_{Al}=\dfrac{5,4}{27}=0,2(mol)\\\Rightarrow n_{HCl}=3n_{Al}=0,6(mol)\\\Rightarrow C_{M_{HCl}}=n/V=\dfrac{0,6}{0,15}=4(M)\\\text{ Câu hỏi 1 : B}\\\Rightarrow n_{AlCl_3}=n_{Al}=0,2(mol)\\\Rightarrow m_{AlCl_3} = 0,2.133,5=26,7(gam)\\\text{ Câu hỏi 2 : A}\\\Rightarrow n_{H_2}=3/2n_{Al}=0,3(mol)\Rightarrow V_{H_2}(đktc)=0,3.22,4=6,72(lít)\\\text{ Câu hỏi 3 : C} \)

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

          0,1-->0,2------>0,1--->0,1

=> \(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

b) m_{dd sau pư} = 5,6 + 200 - 0,1.2 = 205,4 (g)

=> \(C\%=\dfrac{12,7}{205,4}.100\%=6,18\%\)

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\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{HCl}=0,5.1=0,5mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1         0,2          0,1          0,1

\(m_{HCl\left(dư\right)}=\left(0,5-0,2\right).36,5=10,95g\\ b)m_{FeCl_2}=0,1.127=12,7g\\ c)V_{H_2}=0,1.22,4=2,24l\\ d)C_{\%HCl\left(dư\right)}=\dfrac{10,95}{200}\cdot100=5,475\%\\ C_{\%HCl\left(pư\right)}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\)

a) Fe + 2HCl --> FeCl₂ + H₂

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

           0,05<-0,1<-----------0,05

=> m = 0,05.56 = 2,8 (g)

c) \(m_{HCl}=0,1.36,5=3,65\left(g\right)\Rightarrow m_{dd.HCl}=\dfrac{3,65.100}{10}=36,5\left(g\right)\)

`Fe + 2HCl -> FeCl_2 + H_2 \uparrow`

 `0,1`      `0,2`          `0,1`       `0,1`         `(mol)`

`n_[Fe]=[5,6]/56=0,1(mol)`

`a)V_[H_2]=0,1.22,4=2,24(l)`

`b)C_[M_[HCl]]=[0,2]/[0,1]=2(M)`