S + O₂ -> SO₂ (1)

n_S=0,1(mol)

Từ 1:

n_SO2=n_O2=n_S=0,1(mol)

V_O2=22,4.0,1=22,4(lít)

m_SO2=64.0,1=6,4(g)

AgNO₃ + HCl -> AgCl + HNO₃ (1)

n_AgCl=0,1(mol)

Từ 1:

n_HCl=n_AgNO3=0,1(mol)

m_HCl=36,5.0,1=3,65(g)

C% dd HCl=\(\dfrac{3,65}{10}.100\%=36,5\%\)

Fe + S -> FeS (1)

n_Fe=0,1(mol)

n_S=0,2(Mol)

Vì 0,1<0,2 nên sau PƯ 1 còn S dư 0,1 mol

%m_S=\(\dfrac{32.0,1}{12}.100\%=26,67\%\)

%m_FeS=100-26,7=73,33%

a;

Na₂SO₃ + 2HCl -> 2NaCl + SO₂ + H₂O (1)

2Al + 3H₂SO₄ -> Al₂(SO₄)₃ + 3H₂ (2)

n_Na2SO3=0,01(mol)

n_Al=0,01(mol)

THeo PTHH 1 và 2 ta có:

n_Na2SO3=n_SO2=0,01(mol)

n_H2=\(\dfrac{3}{2}\)n_Al=0,015(mol)

VÌ 0,015>0,01 nên n_SO2<n_H2 hay V_SO2 < V_H2

b;

m_{hh khí}=64.0,01+2.0,015=0,67(g)

M_{hh khí}=\(\dfrac{0,67}{0,01+0,015}=26,8\)

VÌ 26,8<29 nên hh khí nhẹ hơn kk

a;

M_A=\(\dfrac{12}{0,25}=48\)

b;

n_B=0,4(mol)

M_B=\(\dfrac{11,2}{0,4}=28\)

CxHyOzNt +(x+\(\dfrac{y}{4}\)-\(\dfrac{z}{2}\) )O2 ----> xCO2 + \(\dfrac{y}{2}\)H2O + \(\dfrac{t}{2}\)N2

c;

2KClO₃ -> 2KCl + 3O₂

S + O₂ -> SO₂

2SO₂ + O₂ -> 2SO₃

SO₃ + H₂O -> H₂SO₄

2Al + 3H₂SO₄ -> Al₂(SO₄)₃ + 3H₂

b;

2KMnO₄ -> K₂MnO₄ + MnO₂ + O₂

C + O₂ -> CO₂

CO₂ + 2NaOH -> Na₂CO₃ + H₂O

Na₂CO₃ + CO₂ + H₂O -> 2NaHCO₃

2NaHCO₃ -> Na₂CO₃ + CO₂ + H₂O

2.

Fe + CuSO₄ -> FeSO₄ + Cu

m_{dd CuSO4}=200.1,12=224(g)

n_CuSO4=\(\dfrac{224.10\%}{160}=0,14\left(mol\right)\)

n_Fe=0,07(mol)

Vì 0,07<0,14 nên sau PƯ CuSO₄ dư 0,07 mol

Theo PTHH ta có:

n_Cu=n_FeSO4=n_Fe=0,07(mol)

m_Cu=64.0,07=4,48(g)

C_M dd FeSO₄=\(\dfrac{0,07}{0,2}=0,35M\)

C_M dd CuSO₄=\(\dfrac{0,07}{0,2}=0,35M\)