Ta có: 0,11>0,12, ta được Fe dư.

a, Theo PT: 

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

Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\n_{HCl}=0,1\cdot1=0,1\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\) \(\Rightarrow\) Sắt còn dư, HCl p/ứ hết

\(\Rightarrow n_{Fe\left(dư\right)}=0,05\left(mol\right)=n_{FeCl_2}\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe\left(dư\right)}=0,05\cdot56=2,8\left(g\right)\\C_{M_{FeCl_2}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\end{matrix}\right.\)

\(a/ \\ Fe+2HCl \rightarrow FeCl_2+H_2\\ n_{Fe}=\frac{5,6}{56}=0,1(mol)\\ n_{HCl}=0,1.1=0,1(mol)\\ Fe: 0,1>HCl: \frac{0,1}{2}=0,05\\ \Rightarrow \text{Fe dư, HCl hết} n_{H_2}=\frac{1}{2}.n_{HCl}=0,05(mol)\\ V_{H_2}=1,12l\\ b/ \\ \text{Fe dư}\\ n_{Fe}=\frac{1}{2}.n_{HCl}=0,05(mol)\\ m_{Fe}=(0,1-0,05).56=2,8g\\ C/ \\ n_{FeCl_2}=\frac{1}{2}.n_{HCl}=0,05(mol)\\ CM_{FeCl_2}=0,5M \)

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

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

\(n_{HCL}=0,1.1=0,1\left(mol\right)\)

\(\Rightarrow n_{H_2}=2nH_2=2,0,1=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(ml\right)\)

b) sau pư Fe dư

ta có 1 molFe Pư 2 molHCL

  0,05 molFe pư 0,1 HCL

\(\Rightarrow n_{Fe\left(dư\right)}:0,1-0,05=0,05\left(mol\right)\)

c)\(C_{MFeCL_2}=\dfrac{2.n_{HCL}}{0,1}=2M\)

a)

$n_{Fe} = \dfrac{11,2}{56} = 0,2(mol) ; n_{HCl} = 0,1.2 = 0,2(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$

Ta thấy : 

$n_{Fe} : 1 > n_{HCl} : 2$ nên Fe dư

$n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,1(mol)$
$V_{H_2} = 0,1.22,4 = 2,24(lít)$

b)

$n_{Fe\ pư} = n_{H_2} = 0,1(mol)$

$\Rightarrow m_{Fe\ dư} = 11,2 - 0,1.56 = 5,6(gam)$

c)

$n_{FeCl_2} = n_{Fe\ pư} = 0,1(mol)$
$C_{M_{FeCl_2}} = \dfrac{0,1}{0,1} = 1M$

a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\)

PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,4}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, - H₂SO₄ dư.

\(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,2\left(mol\right)\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,2.98=19,6\left(g\right)\)

c, \(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

Theo PT: \(n_{FeSO_4}=n_{Fe}=0,2\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(n_{H_2SO_4}=0.2\cdot2=0.4\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(1.........1\)

\(0.2..........0.4\)

\(LTL:\dfrac{0.2}{1}< \dfrac{0.4}{1}\Rightarrow H_2SO_4dư\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{H_2SO_4\left(dư\right)}=\left(0.4-0.2\right)\cdot98=19.6\left(g\right)\)

\(C_{M_{FeSO_4}}=\dfrac{0.2}{0.2}=1\left(M\right)\)

a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(n_{HCl}=0,1.2=0,2\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\), ta được Fe dư.

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, Fe dư.

Theo PT: \(n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow n_{Fe\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe\left(dư\right)}=0,1.56=5,6\left(g\right)\)

c, \(n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow C_{M_{FeCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\)

a) \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right);n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\)

PTHH:         \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Ban đầu:     0,2       0,4

Sau pư:         0        0,2         0,2           0,2

`=> V_{H_2} = 0,2.22,4 = 4,48(l)`

`b) m_{H_2SO_4(dư)} = 0,2.98 = 19,6(g)`

`c)` \(C_{M\left(FeSO_4\right)}=C_{M\left(H_2SO_4.d\text{ư}\right)}=\dfrac{0,2}{0,2}=1M\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\\ Vì:\dfrac{0,2}{1}< \dfrac{0,4}{1}\Rightarrow H_2SO_4dư\\ n_{H_2}=n_{H_2SO_4\left(p.ứ\right)}=n_{FeSO_4}=n_{Fe}=0,2\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b,n_{H_2SO_4\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\\ m_{H_2SO_4}=0,2.98=19,6\left(g\right)\\ c,V_{ddsau}=V_{ddH_2SO_4}=200\left(ml\right)=0,2\left(l\right)\\ C_{MddH_2SO_4\left(dư\right)}=\dfrac{0,2}{0,2}=1\left(M\right)\\ C_{MddFeSO_4}=\dfrac{0,2}{0,2}=1\left(M\right)\)

a. Đổi 200 ml = 0,2 lít

 \(n_{Fe}=\dfrac{11.2}{56}=0,2\left(mol\right)\)

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

PTHH : Fe + 2HCl -> FeCl₂ + H₂

           0,1       0,2          0,1      0,1

Ta thấy : \(\dfrac{0.2}{1}>\dfrac{0.2}{2}\) => Fe dư , HCl đủ

\(m_{Fe\left(dư\right)}=\left(0,2-0,1\right).56=5,6\left(g\right)\)

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

c. Sau phản ứng chất tan là FeCl₂

\(V_{FeCl_2}=0,1.2=0,2\left(l\right)\)

\(\Rightarrow C_{M_{FeCl_2}}=\dfrac{0.1}{0,2}=0,5\left(M\right)\)