\(n_{H2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

a) Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)

            1         2             1          1

          0,15     0,3                      0,15

b) \(n_{Fe}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)

⇒ \(m_{Fe}=0,15.56=8,4\left(g\right)\)

c) \(n_{HCl}=\dfrac{0,15.2}{1}=0,3\left(mol\right)\)

50ml = 0,05l

\(C_{M_{ddHCl}}=\dfrac{0,3}{0,05}=6\left(M\right)\)

 Chúc bạn học tốt

Fe + 2HCl -> FeCl2 + H2
nFe = 5,6/56 = 0,1 mol
=>nH2 = 0,1 mol
=> VH2= 0,1*22,4= 2,24 lít

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

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

          0,1-->0,2------------------>0,1

=> \(\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,2.36,5}{15\%}=\dfrac{146}{3}\left(g\right)\\V_{H_2}=0,1.22,4=4,48\left(l\right)\end{matrix}\right.\)

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

a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

____0,15<--0,3<-------------0,15

=> m_Fe = 0,15.56 = 8,4 (g)

b) \(C_{M\left(ddHCl\right)}=\dfrac{0,3}{0,05}=6M\)

`a)PTHH:`

`Fe + 2HCl -> FeCl_2 + H_2`

`0,4`                      `0,4`       `0,4`

`n_[H_2] = [ 8,96 ] / [ 22,4 ] = 0,4 (mol)`

`b) m_[Fe] = 0,4 . 56 = 22,4 (g)`

`c) m_[FeCl_2] = 0,4 . 127 = 50,8 (g)`

Fe+2HCl->FeCl2+H2

 0.4               0.4      0.4

nH2=V/22.4= 0.4 (mol)

mFe=n*M=0.4*56=22.4(g)

mFeCl2=n*M=0.4*(56+35.5*2)=50.8(g)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

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

\(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)

\(m_{Fe}=0,15.56=8,4\left(g\right)\)

\(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)

\(50ml=0,05l\)

\(C_{M_{ddHCl}}=\dfrac{0,3}{0,05}=6\left(M\right)\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\a, Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,V\text{ì}:\dfrac{0,2}{1}>\dfrac{0,1}{1}\Rightarrow Zn\text{dư}\\ \Rightarrow n_{Zn\left(p.\text{ứ}\right)}=n_{ZnCl_2}=n_{H_2}=0,1\left(mol\right)\\b, m_{Zn\left(p.\text{ứ}\right)}=0,1.65=6,5\left(g\right)\\ n_{HCl}=0,1.2=0,2\left(mol\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ d,m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)

\(a)\\ Fe + 2HCl \to FeCl_2 + H_2\)

b)

\(n_{Fe} = \dfrac{22,4}{56}= 0,4(mol)\\ n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\)

Ta thấy : \(n_{Fe} > n_{H_2}\) nên Fe dư.

Theo PTHH :

\(n_{Fe\ pư} = n_{H_2} = 0,3(mol)\\ \Rightarrow m_{Fe\ pư} = 0,3.56 = 16,8(gam)\)

c)

Ta có : 

\(n_{FeCl_2} = n_{H_2} = 0,3(mol)\\ \Rightarrow m_{FeCl_2} = 0,3.127 = 38,1(gam)\)

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\b.n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\\ c.n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\\ m_{FeCl_2}=0,1.127=12,7\left(g\right) \)