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

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

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)

b, Fe dư.

Theo PT: \(n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\Rightarrow n_{Fe\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)

\(\Rightarrow m_{Fe\left(dư\right)}=0,05.56=2,8\left(g\right)\)

c, \(n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)

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

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

PTHH :

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

trc p/u : 0,1       0,1         

p/u :       0,05    0,1         0,05     0,05 

sau p/u: 0,05      0           0,05       0,05 

---> sau p/ư : Fe dư 

\(a,V_{H_2}=0,05.22,4=1,12\left(l\right)\) 

b, \(m_{Fedư}=0,05.56=2,8\left(g\right)\)

\(c,_{FeCl_2}=0,05.127=6,35\left(g\right)\)

\(m_{ddFeCl_2}=5,6+\left(0,1.36,5\right)-\left(0,05.1\right)=9,2\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{6,35}{9,2}.100\%\approx69\%\)

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

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

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

a, Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)

⇒ V_H2 = 0,05.22,4 = 1,12 (l)

b, Sau pư, Fe dư.

Theo PT: \(n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\Rightarrow n_{Fe\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)

⇒ m_{Fe (dư)} = 0,05.56 = 2,8 (g)

c, Theo PT: n_FeCl2 = n_{Fe (pư)} = 0,05 (mol)

\(\Rightarrow C_{M_{FeCl_2}}=\dfrac{0,05}{0,1}=0,5M\)

Bạn tham khảo nhé!

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

\(n_{HCl}=0.1\cdot1=0.1\left(mol\right)\)

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

\(\dfrac{n_{Fe}}{1}=0.1>\dfrac{n_{HCl}}{2}=\dfrac{0.1}{2}=0.05\)

\(\Rightarrow Fedư\)

\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)

\(m_{Fe\left(dư\right)}=\left(0.1-0.05\right)\cdot56=2.8\left(g\right)\)

\(C_{M_{FeCl_2}}=\dfrac{0.05}{0.1}=0.5\left(M\right)\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right);n_{HCl}=1.0,1=0,1\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,1}{2}< \dfrac{0,1}{1}\Rightarrow Fe.dư\\ n_{H_2}=n_{FeCl_2}=n_{Fe\left(p.ứ\right)}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ b,n_{Fe\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\\ m_{Fe\left(dư\right)}=0,05.56=2,8\left(g\right)\\ c,V_{ddFeCl_2}=V_{ddHCl}=0,1\left(l\right)\\ C_{MddFeCl_2}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)

                  Đổi: 100 ml = 0,1 lít

PTHH:      Fe + 2HCl → FeCl₂ + H₂ ↑

Số mol của HCl là: 0,1 . 1 = 0,1 mol

Số mol của Fe là: 5,6: 56 = 0,1 mol

So sánh: 0,1 > \(\frac{0,1}{2}\) => Fe dư... Tính theo HCl 

Số mol H₂ tạo ra là: 0,1 : 2 = 0,05 mol

Khối lượng H₂ tạo ra là: 0,05 . 2 = 0,1 gam

Khối lượng Fe dư sau pứ là: 0,05 . 56 = 2,8 gam

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

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

LTL: \(0,1>\dfrac{0,1}{2}\) => Fe dư

Theo pthh: \(n_{H_2}=n_{FeCl_2}=n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

=> V_H2 = 0,05.22,4 = 1,12 (l)

b, Chất dư là Fe

m_{Fe (dư)} = (0,1 - 0,05).56 = 2,8 (g)

c, \(C_{M\left(FeCl_2\right)}=\dfrac{0,05}{0,1}=0,5M\)

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

\(n_{HCl}=0.1\cdot1=0.1\left(mol\right)\)

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

\(1............2\)

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

\(LTL:\dfrac{0.1}{1}>\dfrac{0.1}{2}\Rightarrow Fedư\)

\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)

\(m_{Fe\left(dư\right)}=\left(0.1-0.05\right)\cdot56=2.8\left(g\right)\)

\(C_{M_{FeCl_2}}=\dfrac{0.05}{0.1}=0.5\left(M\right)\)

a.
Đổi: 100 ml = 0,1 lít
PTHH: Fe + 2HCl → FeCl2 + H2 ↑
Số mol của HCl là: 0,1 x 1 = 0,1 mol
Số mol của Fe là: 5,6: 56 = 0,1 mol
So sánh: 0,1 > 0,1/2 => Fe dư =>Tính theo HCl
Số mol H2 tạo ra là: 0,1 : 2 = 0,05 mol
=> VH2 = 0,05.22,4= 1,12l
b. Fe dư
nFe dư = 0,1 - 0,05 = 0,05 mol
Khối lượng Fe dư sau pứ là: 0,05 . 56 = 2,8 gam
c.
nFeCl2 = 1/2.nHCl = 0,1/2=0,05 mol
CM FeCl2 = 0,05/0,1 = 0,5M

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)

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