Sắt tác dụng được với: HCl, CuSO₄, Cl₂.

a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

Chất rắn còn lại sau pư là Cu.

b, Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)

⇒ m_Zn = 0,1.65 = 6,5 (g)

⇒ m _{chất rắn} = 10,5 - 6,5 = 4 (g)

Câu 5.1

\(n_{H_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

_____0,4_____0,8__________0,4 (mol)

a, \(m_{Fe}=0,4.56=22,4\left(g\right)\)

b, \(C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)

Câu 5.2

\(n_{H_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

_____0,4_____0,8___________0,4 (mol)

a, \(m_{Mg}=0,4.24=9,6\left(g\right)\)

b, \(C_{M_{HCl}}=\dfrac{0,8}{0,4}=2\left(M\right)\)

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

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.24,79=2,479\left(l\right)\)

Có lẽ đề cho 200ml H₂SO₄ chứ không phải 200 mol bạn nhỉ?

Ta có: \(n_{Al}=\dfrac{21,6}{27}=0,8\left(mol\right)\)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

a, Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=1,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=1,2.24,79=29,748\left(l\right)\)

b, \(n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=1,2\left(mol\right)\)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{1,2}{0,2}=6\left(M\right)\)

Ta có: \(n_{H_2}=\dfrac{65}{2}=32,5\left(mol\right)\)

PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Theo PT: \(n_{H_2O}=n_{H_2}=32,5\left(mol\right)\)

\(\Rightarrow m_{H_2O}=32,5.18=585\left(g\right)\)

Ta có: n_HCl = 2n_H2 = 1 (mol)

Theo ĐLBT KL, có: m_KL + m_HCl = m _muối + m_H2

⇒ m _muối = 13,4 + 1.36,5 - 0,5.2 = 48,9 (g)

1. \(BaO+N_2O_5\rightarrow Ba\left(NO_3\right)_2\)

2. \(Na_2O+2HCl\rightarrow2NaCl+H_2O\)

3. \(CaO+2HCl\rightarrow CaCl_2+H_2O\)

4. \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

5. \(2Na+Cl_2\underrightarrow{t^o}2NaCl\)

6. \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

7. \(Al\left(OH\right)_3+3HCl\rightarrow AlCl_3+3H_2O\)

8. \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

9. \(2K+Cl_2\underrightarrow{t^o}2KCl\)

10. \(2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\)

Câu 1:

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b, Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

Theo PT: \(n_{MgCl_2}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)

c, \(m_{MgCl_2}=0,2.95=19\left(g\right)\)

d, \(n_{HCl}=2n_{Mg}=0,4\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)

Câu 2:

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, Ta có: \(n_{FeCl_2}=\dfrac{19,05}{127}=0,15\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=n_{FeCl_2}=0,15\left(mol\right)\)

⇒ m_Fe = 0,15.56 = 8,4 (g)

V_H2 = 0,15.24,79 = 3,7185 (l)