\(n_{HCl}=0,1.0,2=0,02\left(mol\right)\\ n_{H_2SO_4}=0,1.0,1=0,01\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ n_{NaOH}=n_{HCl}+2.n_{H_2SO_4}=0,04\left(mol\right)\\ V_{\text{dd}NaOH}=\dfrac{0,04}{0,5}=0,08\left(l\right)\\ \Rightarrow Ch\text{ọn}.A\)

\(CH_2\)

n(CH2=CH2)-to\Xt->-(-CH2-CH2-)-_n

39: B

40: A

41: B

a) \(n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Có \(\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,4}{0,2}=2\) => Tạo ra muối CO₃^2-

PTHH: 2NaOH + CO₂ --> Na₂CO₃ + H₂O

________0,4--->0,2-------->0,2

=> m_Na2CO3 = 0,2.106 = 21,2 (g)

b) \(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

n_Ca(OH)2 = 0,2.0,2 = 0,04 (mol)

PTHH: Ca(OH)₂ + CO₂ --> CaCO₃\(\downarrow\) + H₂O

_______0,04--->0,04------->0,04

CaCO₃ + CO₂ + H₂O --> Ca(HCO₃)₂

_0,01<---0,01-------------->0,01

=> m_CaCO3 = (0,04-0,01).100 = 3(g)

=> m_Ca(HCO3)2 = 0,01.162 = 1,62 (g)

c) \(n_{CO_2}=\dfrac{4,4}{44}=0,1\left(mol\right)\)

n_KOH = 0,15.1 = 0,15 (mol)

PTHH: 2KOH + CO₂ --> K₂CO₃ + H₂O

______0,15-->0,075---->0,075

K₂CO₃ + CO₂ + H₂O --> 2KHCO₃

0,025<-0,025------------->0,05

=> m_K2CO3 = (0,075-0,025).138 = 6,9 (g)

=> m_KHCO3 = 0,05.100 = 5(g)

C2H4+Br2-to>C2H4Br2

0.6----0,6

n Br2=\(\dfrac{96}{160}\)=0,6 mol

m CH4=24,8-0,6.28=8g

=>n CH4=\(\dfrac{8}{16}\)=0,5 mol

=>%CH4=\(\dfrac{0,5.22,4}{0,5.22,4+0,6.22,4}\).100=45,45%

=>%C2H4=54,55%

mCu=0,3*3/2 * 64=28,8(g)

m muối = 0,45*188=84,6(g)