a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

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

c, Theo PT: \(n_{HCl}=2n_{Fe}=0,4\left(mol\right)\)

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

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

\(1mol\)     \(2mol\)                 \(1mol\)

\(0,2mol\)    \(0,4mol\)          \(0,2mol\)

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

\(V_{H_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)

\(m_{HCl}=n.M=0,4.=14,6\left(g\right)\)

a) PTHH:

Zn + HCl - - -> ZnCl₂ + H₂

Zn + 2HCl -----> ZnCl₂ + H₂

b) Áp dụng định luật bảo toàn khối lượng, ta có:

m_Zn + m_HCl = m_ZnCl2 + m_H2

m_H2 = 3,25 + 3,65 - 6,8 = 0,1 (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) => \(2n_{Fe}=n_{HCl}=0.4\left(mol\right)\)

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

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\\ n_{H_2}=n_{Fe}=0,25\left(mol\right)\\ V_{H_2}=0,25.22,4=5,6\left(l\right)\\ n_{HCl}=2n_{Fe}=0,5\left(mol\right)\\ m_{HCl}=0,5.36,5=18,25\left(g\right)\)

1. \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

2. \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

3. Ta có: \(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)

PT: \(Fe_3O_4+4H_2\underrightarrow{^{t^o}}3Fe+4H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,2}{4}\), ta được Fe₃O₄ dư.

Theo PT: \(n_{Fe_3O_4\left(pư\right)}=\dfrac{1}{4}n_{H_2}=0,05\left(mol\right)\Rightarrow n_{Fe_3O_4\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4\left(dư\right)}=0,05.232=11,6\left(g\right)\)

1. \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

2. \(n_{zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{13}{65}=0,2\left(mol\right)\)

Theo PTHH: \(n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\)

\(\Rightarrow m_{ZnCl_2}=n_{ZnCl_2}.M_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

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

3. \(2H_2+Fe_3O_4\rightarrow3Fe+2H_2O\)

2 mol------1 mol------3 mol--2 mol

\(n_{Fe_3O_4}=\dfrac{m_{Fe_3O_4}}{M_{Fe_3O_4}}=\dfrac{23,2}{232}=0,1\left(mol\right)\)

\(\dfrac{n_{Fe_3O_4}}{1}=\dfrac{0,1}{1}\)

\(\dfrac{n_{H_2}}{2}=\dfrac{0,2}{2}\)

\(\dfrac{n_{Fe_3O_4}}{1}=\dfrac{n_{H_2}}{2}\)

Vậy không có chất nào dư cả

a) PTHH: Fe + 2HCl --> FeCl₂ + H₂

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

            0,2-->0,4------>0,2-->0,2

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

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

d) \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

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