a) Pt: $Fe+2HCl\to FeC{l}_2+H_2$

b) nFe = \(\dfrac{11,2}{56}=0,2mol\)

Theo pt: nH₂ = nFe = 0,2 mol

=> VH₂ = 0,2.22,4 = 4,48lit

c) Theo pt: nHCl = 2nFe = 0,4 mol

=> mHCl = 0,4.36,5 = 14,6 g

=> \(C\%=\dfrac{14,6}{73}.100\%=20\%\)

\(a) Fe + 2HCl \to FeCl_2 + H_2\\ b) n_{FeCl_2} = n_{Fe} =\dfrac{11,2}{56} = 0,2(mol)\\ m_{FeCl_2} = 0,2.127 = 25,4(gam)\\ c) n_{H_2} = n_{Fe} = 0,2(mol)\Rightarrow V_{H_2} = 0,2.22,4 = 4,48(lít)\\ d) n_{HCl} = 2n_{Fe} = 0,4(mol)\\ C\%_{HCl} = \dfrac{0,4.36,5}{300}.100\% = 4,867\%\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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. \(n_{Zn}=\dfrac{6.5}{65}=0,1\left(mol\right)\)

PTHH : Zn + 2HCl -> ZnCl₂ + H₂

            0,1      0,2                  0,1

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

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

\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,1      0,2                       0,1

\(V_{H_2}=0,1\cdot22,4=2,24l\)

\(m_{HCl}=0,2\cdot36,5=7,3g\)

a)

$Fe + 2HCl \to FeCl_2 + H_2$
b)

Theo PTHH : 

$n_{FeCl_2} = n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$m_{FeCl_2} = 0,2.127 = 25,4(gam)$

a) Có khí thoát ra

Fe + 2HCl --> FeCl₂ + H₂

b) 

Theo ĐLBTKL: m_Fe + m_HCl = m_FeCl2 + m_H2

=> m_HCl = 25,4 + 0,4 - 11,2 = 14,6(g)

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)