Решение задач по химии "Alkanes" 8-11 класс

Alkanes
1. Calculate of what mass of carbon in 90 g of ethane C
2
H
6
.
given:
m
2
Н
6
) = 90 g
find:
m(С) - ?
Solution:
М
2
Н
6
) = 30 g/mol
90 g С
2
Н
6
------ х g С
30 g С
2
Н
6
----- 24 g С
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

Answer: m(С) = 72 g
2. The thermal decomposition 100 liters (under normal conditions) of methane
formed 43.4 g of carbon - sazhi. Determine the output of soot from
theoretically possible.
given:
V(СН
4
) = 100 liters
m
teor.
(С) = 43,4 g
find:
η(С) -?
Solution:
СН
4
→ С + 2Н
2
М(С) = 12 g/mol
V
m
= 22,4 lit/mol
100 lit СН
4
------ х g С
22,4 lit СН
4
------- 12 g С
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


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

Answer: η = 0,810.
3. Having a thermochemical equation methane combustion СН
4(g)
+ 2О
2(g)
=
СО
2(g)
+ Н
2
О
(g)
+ 800 kJ, under normal conditions (under normal conditions)
determine the amount of methane which should be burned to produce 10 000
kJ heat.
given:
Q = 10000 kJ
find:
V(СН
4
) -?
Solution:
СН
4(g)
+ 2О
2(g)
= СО
2(g)
+ Н
2
О
(g)
+ 800 kJ
V
m
= 22,4 lit/mol
х lit СН
4
------ 10000 kJ
22,4 lit СН
4
------- 800 kJ
  


Answer: V(СН
4
) = 280 lit
4. Having a thermochemical equation methane combustion СН
4(g)
+ 2О
2(g)
=
СО
2(g)
+ Н
2
О
(g)
+ 800 kJ, determine the amount of heat generated by the
combustion of 100 g of methane.
given:
m(СН
4
) = 100 g
find:
Q -?
Solution:
СН
4(g)
+ 2О
2(g)
= СО
2(g)
+ Н
2
О
(g)
+ 800 kJ
М(СН
4
) = 16 g/mol
100 g СН
4
------ х kJ
16 g СН
4
------- 800 kJ
  


Answer: Q = 5000 kJ
5. Calculate of attitude mass elements in methane (CH
4
) .
given:
СН
4
find:
m(С) : m(Н) -?
Solution:
1) Мr(СН
4
) = Аr(С) + Аr(Н);
Мr(СН
4
) = 12 + 4 = 16.
2) m(С) : m(Н) = 12 : 4

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

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



Answer: m(С) : m(Н) = 3 : 1.
6. Install formula substances maccovaya share of carbon in which 82.75 %
and 17.25 % hydrogen . Relative density vapor hydrogen substance is equal
to 29 .
given:
ω(С) = 82,75 %
ω(Н) = 17,25 %
Dн
2
х
Н
у
) = 29
find:
С
х
Н
у
Solution:
М
х
Н
у
) = 2D
H2
х
Н
у
);
М
х
Н
у
) = 2 g/mol · 29 = 58 g/mol.
m
х
Н
у
) = М
х
Н
у
) · ν(С
х
Н
у
);
m(С
х
Н
у
) = 58 g/mol · 1 mol = 58 g.
m(С) = ω(С) · m
х
Н
у
); m(С) = 0,8275 · 58 g = 48 g;
m(Н) = ω(Н) · m
х
Н
у
); m(Н) = 0,1725 · 58 g = 10 g.

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




ν(С) : ν(Н) = 4 : 10 formula of the substance C
4
H
10
- butane
Answer: formula of the substance C
4
H
10
.
7. At the combustion of 0.57 g of an organic matter in excess of oxygen was
formed 1.76 g of carbon oxide (IV) and 0.81 g of water. Byvedite molecular
formula of the substance , if the relative density ego vapor in the air is equal
to 3,931 .
given:
m(substance) = 0,57 g
m(СО
2
) = 1,76 g
m(Н
2
О) = 0,81 g
D
air
(substance) = 3,931
find: formula of the substance
Solution:
M(substance) = 29D
air
(substance);
М(substance) = 29 g/mol · 3,931 = 114 g/mol.
М
2
О) = 18 g/mol
М(СО
2
) = 44 g/mol








ν(С) = ν(СО
2
) = 0,04 mol








ν(Н) = 2ν(Н
2
О) ; ν(Н) = 2 · 0,045 = 0,09 mol
m(С) = ν(С) · М(С);
m(С) = 0,04 mol · 12 g/mol = 0,48 g;
m(Н) = ν(Н) · М(Н);
m(Н) = 0,09 mol · 1 g/mol = 0,09 g.
m(С) + m(Н) = 0,48 g + 0,09 g = 0,57 g.
ν(С) : ν(Н) = 0,04 : 0,09 = 4 : 9.
The of simplest formula of substance C
4
H
9
;
M(C
4
H
9
) = 57 g/mol (less of in true).

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

 
Answer: Molecular formula of the substance С
8
H
18
.
8. Determine the molecular formula of a halogen , if the mass fraction of
carbon in it is 24% , the mass fraction of chlorine - 70% , and the relative
density of the vapor in the air is 1.74 .
given:
ω(С) = 24 %
ω(Сl) = 70 %
D
air
. = 1,74
find:
С
х
Н
у
Сl
z
Solution:





Mr(С
х
Н
у
Сl
z
) = D
(air)
· Mr
(air)
= 29 · 1,47 = 50,5.
ω(Н) = 100 % - 24 % - 70 % = 6 %.

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
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



The simplest formula CH
3
Сl
Mr(CH
3
Сl) = 12 + 1∙3 + 35,5 = 50,5
Answer: CH
3
Сl
9. Install the molecular formula of the substance in which the mass fraction of
80% carbon and hydrogen mass fraction of 20%, and the molar mass is 30
g/mol.
given:
ω(С) = 80 %
ω(Н) = 20 %
m
х
Н
у
) = 30 g/mol
find:
С
х
Н
у
Solution:






 





The simplest formula СН
3
; Mr(СH
3
) = 15




M(C
2
H
6
) = 12∙ 2+ 1∙ 6 = 30 g/mol
Answer: C
2
H
6
(ethane)
10. Calculate the volume methane of given to normal conditions, which can
be obtained by heating acetic acid weighing 24 grams with an excess of
sodium hydroxide. Mass fraction of methane is of 35 %.
given:
m(CH
3
СООН) = 24 g
ω
в
(СН
4
) = 35 %;
find:
V
praс.
(СН
4
) -?
Solution:






3
СООН + 2NаОН→ СН
4
+ Nа
2
СО
3
+ Н
2
О
ν(СН
4
) = ν(CН
3
СООН); ν(СН
4
) = 0,4 mol

 


  


V
m
= 22,4 l/mol
V
.
(СН
4
) =ν(СН
4
) · V
m
;
V(СН
4
) =0,14 mol · 22,4 l/mol = 3,14 lit
Answer: V(СН
4
) = 3,14 lit