a) Na₂SO₃ + 2HCl --> 2NaCl + SO₂ + H₂O

b) n_HCl = 0,2.1 = 0,2 (mol)

Na₂SO₃ + 2HCl --> 2NaCl + SO₂ + H₂O

_0,1<------0,2------->0,2----->0,1

m_NaCl = 0,2.58,5 = 11,7(g)

V_SO2 = 0,1.22,4 = 2,24 (l)

c) m_Na2SO3 = 0,1.126 = 12,6 (g)

d) \(C_{M\left(NaCl\right)}=\dfrac{0,2}{0,2}=1M\)

a) \(n_R=\dfrac{5,85}{M_R}\left(mol\right)\)

\(n_{H_2SO_4}=0,15.0,5=0,075\left(mol\right)\)

PTHH: 2R + 2H₂O --> 2ROH + H₂

____\(\dfrac{5,85}{M_R}\)------------>\(\dfrac{5,85}{M_R}\)

2ROH + H₂SO₄ --> R₂SO₄ + H₂O

0,15<-0,075

=> \(\dfrac{5,85}{M_R}=0,15=>M_R=39\left(g/mol\right)\) => R là K(Kali)

b) 

\(n_K=\dfrac{5,85}{39}=0,15\left(mol\right)\)

PTHH: 2K + 2H₂O --> 2KOH + H₂

_____0,15---------------------->0,075

=> V_H2 = 0,075.22,4 = 1,68(l)

c) 

PTHH: 4K + O₂ --t^o--> 2K₂O

_____0,15------------>0,075

=> m_K2O = 0,075.94 = 7,05(g)

1) 2Fe + 6H₂SO₄ --> Fe₂(SO₄)₃ + 3SO₂ + 6H₂O

2) 4Mg + 5H₂SO₄ --> 4MgSO₄ + H₂S + 4H₂O

3) 8Al + 30HNO₃ --> 8Al(NO₃)₃ + 3NH₄NO₃ + 9H₂O

4) Mg + 4HNO₃ --> Mg(NO₃)₂ + 2NO₂ + 2H₂O

5) 3Zn + 8HNO₃ --> 3Zn(NO₃)₂ + 2NO + 4H₂O

6) 2P + 5H₂SO₄ --> 2H₃PO₄ + 5SO₂ + 2H₂O

7) S + 2HNO₃ --> H₂SO₄ + 2NO

8) 3H₂S + 4HClO₃ --> 4HCl + 3H₂SO₄

9) 8Fe + 30HNO₃ --> 8Fe(NO₃)₃ + 3N₂O + 15H₂O

10) 5Cu + 12HNO₃ --> 5Cu(NO₃)₂ + N₂ + 6H₂O

trời ơi cứu tinh của tui 

Gọi số mol Cl₂, O₂ là a, b

=> a+b = \(\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Có 71a + 32b = 19,7 - 7,8 = 11,9

=> a = 0,1 ; b = 0,15

Gọi số mol Mg, Al là x, y

=> 24x + 27y = 7,8

Mg⁰ - 2e --> Mg⁺²

x---->2x

Al⁰ - 3e --> Al⁺³

y--->3y

Cl₂⁰ +2e--> 2Cl^-

0,1-->0,2

O₂⁰ +4e--> 2O^-2

0,15->0,6

Bảo toàn e: 2x + 3y = 0,8 

=> x = 0,1 ; y = 0,2

\(\%Al=\dfrac{0,2.27}{7,8}.100\%=69,23\%\) 

=> B

8Zn + 10H₂SO₄ --> 8ZnSO₄ + 2H₂S + 8H₂O

Số phân tử H₂SO₄ bị khử = 2

Số phân tử H₂SO₄ tham gia phản ứng = 10

=> x = 1/5

\(n_M=\dfrac{1,6}{M_M}\left(mol\right)\)

PTHH: M + Cl₂ --t^o--> MCl₂

____\(\dfrac{1,6}{M_M}\)----------->\(\dfrac{1,6}{M_M}\)

=> \(\dfrac{1,6}{M_M}\left(M_M+71\right)=4,44=>M_M=40\left(Ca\right)\)

\(n_{Ca}=\dfrac{1,6}{40}=0,04\left(mol\right)\)

PTHH: Ca + Cl₂ --t^o--> CaCl₂

_____0,04->0,04

=> V_Cl2 = 0,04.22,4 = 0,896(l)

cám ơn ạ !!!

30C

31: A

32: C

33: C

34: A

35: A

36: D

37: C

38: A

39: A

40: D

a)

\(n_{HCl}=\dfrac{200.14,6}{100.36,5}=0,8\left(mol\right)\)

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: MgCO₃ + 2HCl --> MgCl₂ + CO₂ + H₂O

______a--------->2a-------->a-------->a

CaCO₃ + 2HCl --> CaCl₂ + CO₂ + H₂O

_b-------->2b-------->b------->b

=> \(\left\{{}\begin{matrix}84a+100b=28,4\\a+b=0,3\end{matrix}\right.=>\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%MgCO_3=\dfrac{0,1.84}{28,4}.100\%=29,577\%\\\%CaCO_3=\dfrac{0,2.100}{28,4}.100\%=70,423\%\end{matrix}\right.\)

b) m_{dd sau pư} =  28,4 + 200 - 0,3.44 = 215,2 (g)

\(\left\{{}\begin{matrix}C\%\left(MgCl_2\right)=\dfrac{0,1.95}{215,2}.100\%=4,4\%\\C\%\left(CaCl_2\right)=\dfrac{0,2.111}{215,2}.100\%=10,32\%\\C\%\left(HCl\right)=\dfrac{\left(0,8-2.0,1-2.0,2\right).36,5}{215,2}.100\%=3,39\%\end{matrix}\right.\)

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