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 - Chemistry

प्रश्न

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

$C_{3} H_{8} + 5 O_{2} ⟶ 3 CO_{2} + 4 H_{2} O$

$2 C_{4} H_{10} + 13 O_{2} ⟶ 8 CO_{2} + 10 H_{2} O$

संख्यात्मक

उत्तर

Ratio of propane and Butane in mixture

= 60 : 40

= 3 : 2

Ratio in 10 litres mixture = $\frac{3}{5} \times 10 and \frac{2}{5} \times 10$

= 6 lit : 4 lit

$C_{3} H_{8} + 5 O_{2} ⟶ 3 CO_{2} + 4 H_{2} O$
1 vol 3 vol

1 volume of propane produces CO₂ = 3 volumes

∴ 6 litres produces CO₂ = 3 x 6

= 18 litres ...(i)

$2 C_{4} H_{10} + 13 O_{2} ⟶ 8 CO_{2} + 10 H_{2} O$
2 vol 8 vol

2 litres of Butane produces CO₂ = 8 litres

∴ 4 litres produce CO₂ = $\frac{8 \times 4}{2}$

= 16 litres ...(ii)

∴ CO₂ added to atmosphere = 18 + 16

= 34 litres.

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Fundamental Laws of Gases - Gay Lussac’s Law of Combining Volumes

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 17 Q 16 Q 18

पाठ 5: Mole concept and Stoichiometry - Exercise 5A [पृष्ठ ७५]

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सेलिना Concise Chemistry [English] Class 10 ICSE

पाठ 5 Mole concept and Stoichiometry
Exercise 5A | Q 17 | पृष्ठ ७५

संबंधित प्रश्‍न

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$C_{3} H_{8} + 5 O_{2} ⟶ 3 CO_{2} + 4 H_{2} O$

What volume of propane is consumed on using 1000 cm³of air, considering only 20% of air contains oxygen?

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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 - Chemistry

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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 - Chemistry

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