Sai:

NaO₂ -> Na₂O

Mg₂O -> MgO

K₂PO₄ -> K₃PO₄

Ca₂(PO₄)₃ -> Ca₃(PO₄)₂

AgCl₂ -> AgCl

Mg₂CO₃ -> MgCO₃

Đúng:

Fe₂O₃

KNO₂

NaHCO₃

n_H2=\(\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

n_HCl=2n_H2=0,4(mol)

n_HCl=n_Cl=0,4(mol)

Theo ĐLBTKL cho cả bài ta có:

m_Al,Fe + m_Cl=m_a

=>m_a=5,5+35,5.0,4=19,7(g)

Fe + S -> FeS

Vì \(\dfrac{a}{56}< \dfrac{a}{32}\) nên S dư

S + O₂ -> SO₂

2Al + 2NaOH + 2H₂O -> 2NaAlO₂ + 3H₂ (1)

2KMnO₄ + 16HCl -> 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O (2)

2KClO₃ -> 2KCl + 3O₂ (3)

2H₂ + O₂ -> 2H₂O (4)

H₂ + Cl₂ -> 2HCl (5)

n_Al=\(\dfrac{5,94}{27}=0,22\left(mol\right)\)

n_KMnO4=\(\dfrac{1,896}{158}=0,012\left(mol\right)\)

n_KClO3=\(\dfrac{12,25}{122,5}=0,1\left(mol\right)\)

Fe + 2 AgNO₃ -> Fe(NO₃)₂ + 2Ag

n_Fe=\(\dfrac{11,2}{56}=0,2\left(mol\right)\)

n_AgNO3=\(\dfrac{400.17\%}{170}=0,4\left(mol\right)\)

Vì 0,2.2=0,4 nên PƯ xảy ra vừa đủ

TheoPTHH ta có:

n_Fe=n_Fe(NO3)2=0,2(mol)

n_AgNO3=n_Ag=0,4(mol)

m_Fe(NO3)2=0,2.180=36(g)

m_dd =11,2+400-108.0,4=368(g)

C_% dd Fe(NO₃)₂=\(\dfrac{36}{368}.100\%=9,78\%\)

m_3Ca=40.3.1,6605.10^-24=199,26.10^-24(g)

m_7Ba=137.7.1,6605.10^-24=1592,4195.10^-24(g)

m_3Zn=3.65.1,6605.10^-24=323,7975.10^-24(g)

m_2Mg=1,6605.10^-24.2.24=79,704.10^-24 (g)

m_5O=1,6605.10^-24.5.16=132,84.10^-24(g)

m_4Fe=1,6605.10^-24.4.56=371,952.10^-24(g)

2H₂ + O₂ -> 2H₂O

n_H2=\(\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

n_O2=\(\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Vì \(\dfrac{0,25}{2}< \dfrac{0,3}{1}\) nên O₂ dư 0,175(mol)

V_{O2 dư}=0,175.22,4=3,92(lít)

Zn + 2HCl -> ZnCl₂ + H₂

n_Zn=\(\dfrac{6,5}{65}=0,1\left(mol\right)\)

Theo PTHH ta có:

n_HCl=2n_Zn=0,2(mol)

n_Zn=n_ZnCl2=0,1(mol)

m_HCl=0,2.36,5=7,3(g)

m_ZnCl2=136.0,1=13,6(g)

Cho quỳ tím vào 3 dd trên nhận ra:

+NaOH làm quỳ tím chuyển xanh

+HCl là quỳ tím chuyển đỏ

+AlCl₃ ko làm đổi màu quỳ tím