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LPG stands for liquefied petroleum gas. Varieties of LPG are marketed including a mixture of propane (60%) and butane (40%). If 10 litres of this mixture is burnt, - Chemistry

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प्रश्न

LPG stands for liquefied petroleum gas. Varieties of LPG are marketed including a mixture of propane (60%) and butane (40%). If 10 litres of this mixture is burnt, find the total volume of carbon dioxide gas added to the atmosphere. Combustion reaction can be represented as:

\[\ce{C3H8_{(g)} + 5O2_{(g)} -> 3CO2_{(g)} + 4H2O_{(g)}}\]

\[\ce{2C4H10_{(g)} + 13O2_{(g)} -> 8CO2_{(g)} + 10H2O_{(g)}}\]

संख्यात्मक

उत्तर

% of propane = 60%

% of butane = 40%

Total volume of the mixture = 10 litres

Propane = `60/100 xx 10` = 6 litres

Butane = `40/100 xx 10` = 4 litres

(i) Combustion of propane:

C3H8(g) 5O2(g) 3CO2(g) 4H2O(g)
1 vol.           3 vol.
6 litres           ?

1 vol of propane releases 3 vol of CO2

6 litres of propane releases `3/1 xx 6` litres

(ii) Combustion of butane:

2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(g)
2 vol.       8 vol.    
1 vol.       4 vol.    
4 litres       ?    

2 vol of butane releases 8 vol of CO2

4 litres of butane releases `8/2 xx 4` = 16 litres

Total COreleased O2 = 18 L + 16 L

= 34 litres of COgas is added to the atmosphere.

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Numerical Problems of Chemical Equation
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Mole Concept And Stoichiometry - Exercise 9 [पृष्ठ १२३]

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फ्रैंक Chemistry - Part 2 [English] Class 10 ICSE
अध्याय 5 Mole Concept And Stoichiometry
Exercise 9 | Q 5.1 | पृष्ठ १२३

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