Bài 1:

1) Fe + 2HCl --> FeCl₂ + H₂

2) \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂
_____0,3--------------->0,3--->0,3

=> n_H2 = 0,3.22,4 = 6,72(l)

3) m_FeCl2 = 0,3.127=38,1(g)

Bài 2

1) 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

2) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

______0,2<----------------------------------0,3

=> m_Al = 0,2.27 = 5,4(g)

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

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:     0,2     0,4                   0,2

b, \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c, \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

Fe+2HCl→ H₂+FeCl₂

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

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

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

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

\(m_{HCl}=0.4\cdot36.5=14.6\left(g\right)\)

PTHH :    \(Fe+2HCl-->FeCl_2+H_2\uparrow\)  (1)

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

Từ (1) => \(n_{Fe}=n_{H_2}=0.2\left(mol\right)\)

=> \(V_{H2\left(đktc\right)}=n.22,4=0,2.22,4=4,48\left(l\right)\)

Từ (1) => \(2n_{Fe}=n_{HCl}=0.4\left(mol\right)\)

=> \(m_{HCl}=n.M=0,4.\left(1+35.5\right)=14.6\left(g\right)\)

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<--0,15

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

=> \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)

b) \(m_{FeCl_2}=0,15.127=19,05\left(g\right)\)

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

\(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\Rightarrow m_{HCl}=0,3.36,5=10,95\left(g\right)\)

b, \(n_{FeCl_2}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{FeCl_2}=0,15.127=19,05\left(g\right)\)

\(a/n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,15       0,3           0,15          0,15

\(m_{Fe}=0,15.56=8,4g\\ m_{HCl}=0,3.36,5=10,95g\\ b/m_{FeCl_2}=0,15.127=19,05g\)

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

0,15   0,3          0,15       0,15

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

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

\(m_{HCl}=0,3.36,5=10,95\left(g\right)\)

\(b,m_{FeCl_2}=0,15.127=19,05\left(g\right)\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2(mol)\\ a,Fe+2HCl\to FeCl_2+H_2\\ b,n_{FeCl_2}=n_{H_2}=n_{Fe}=0,2(mol)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48(l)\\ m_{FeCl_2}=0,2.127=25,4(g)\)

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

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

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

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

\(m_{FeCl_2}=0.2\cdot127=25.4\left(g\right)\)

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

b. số mol của 16,8 gam Fe:

\(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

Khối lượng của HCl:

\(m_{HCl}=n.M=0,6.36,5=21,9\left(g\right)\)

c.Thể tích khí Hiđro (đktc):
\(V_{H_2}=n.22,4=0,3.22,4=6,72\left(l\right)\)

a, nFe = 11,2/56 = 0,2 (mol)

b, PTHH: Fe + 2HCl -> FeCl2 + H2

Mol: 0,2 ---> 0,4 ---> 0,2 ---> 0,2

mFeCl2 = 0,2 . 127 = 25,4 (g)

c, VH2 = 0,2 . 22,4 = 4,48 (l)

d, CMddHCl = 0,4/0,4 = 1M

`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)