English

LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to atmosphere. CA3HA8+5OA2⟶3COA2+4HA2O 2CA4HA10+13OA2⟶8COA2+10HA2O

Advertisements
Advertisements

Question

LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to atmosphere.

\[\ce{C3H8 + 5O2 → 3CO2 + 4H2O}\]

\[\ce{2C4H10 + 13O2 → 8CO2 + 10H2O}\]

Numerical
Advertisements

Solution

Ratio of propane and Butane in mixture 

= 60 : 40

= 3 : 2

Ratio in 10 litres mixture = `3/5 xx 10 and 2/5 xx 10`

= 6 lit : 4 lit

\[\ce{C3H8 + 5O2 → 3CO2 + 4H2O}\]
1 vol                       3 vol

1 volume of propane produces CO2 = 3 volumes

∴ 6 litres produces CO2 = 3 x 6

= 18 litres   ...(i)

\[\ce{2C4H10 + 13O2 → 8CO2 + 10H2O}\]
2 vol                              8 vol

2 litres of Butane produces CO2 = 8 litres

∴ 4 litres produce CO2 = `(8 xx 4)/2`

= 16 litres   ...(ii)

∴ CO2 added to atmosphere = 18 + 16

= 34 litres.

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Mole concept and Stoichiometry - EXERCISE-5A [Page 75]

APPEARS IN

S.P. Singh Concise Chemistry [English] Class 10 ICSE
Chapter 5 Mole concept and Stoichiometry
EXERCISE-5A | Q 17. | Page 75

RELATED QUESTIONS

State Gay-Lussac’s law of combining volumes.


How does Avogadro's law explain Gay - lussac's law of combining volumes?


Calcium carbide is used for the artificial ripening of fruits. Actually the fruit ripens because of the heat evolved while calcium carbide reacts with the moisture. During this reaction calcium hydroxide and acetylene gas are formed. If 200 cm3 of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below.

\[\ce{2C2H2_{(g)} + 5O2_{(g)}-> 4CO2_{(g)} + 2H2O_{(g)}}\]


24 cc Marsh gas (CH4) was mixed with 106 cc oxygen and then exploded. On cooling the volume of the mixture became 82 cc, of which, 58 cc was unchanged oxygen. Which law does this experiment support? Explain with calculations.


What volume of propane is burnt for every 500 cm3 of air used in the reaction under the same conditions? (assuming oxygen is `1/5`th of air)

\[\ce{C3H8 + 5O2 → 3CO2 + 4H2O}\]


450 cm3 of nitrogen monoxide and 200 cm3 of oxygen are mixed together and ignited. Caclulate the composition of resulting mixture.

\[\ce{2NO + O2 → 2NO2}\]


If 6 liters of hydrogen and 4 liters of chlorine are mixed and exploded and if water is added to the gases formed, find the volume of the residual gas.


112 cm3 of H2S(g) is mixed with 120 cm3 of Cl2(g) at STP to produce HCl(g) and sulphur(s). Write a balanced equation for this reaction and calculate

  1. the volume of gaseous product formed.
  2. composition of the resulting mixture.

112 cm3 of H2S(g) is mixed with 120 cm3 of Cl2(g) at STP to produce HCl(g) and sulphur(s). Calculate the volume of gaseous product formed.


112 cm3 of H2S(g) is mixed with 120 cm3 of Cl2(g) at STP to produce HCl(g) and sulphur(s). Calculate composition of the resulting mixture.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×