Question

What volume of oxygen is required to burn completely 90 dm3 of butane under similar conditions of temperature and pressure?
2C4H10 + 13O2 ⟶ 8CO2 + 10H2O

August 2, 2024

Answer

[By Lussac’s law]

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

2 vol : 13 vol ⟶ 8 vol : 10 vol

To calculate the volume of oxygen

C₄H₁₀ : O₂

2 : 13

90 : x

⇒ x = $13\over 2$ × 90 = 585 dm³

Hence, the volume of oxygen is required is 585 dm³

💡 Some Related Questions

Copper [II] sulphate soln. reacts with sodium hydroxide soln. to form copper hydroxide according to the equation :
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What mass of copper hydroxide is precipitated by using 200 g of sodium hydroxide.
[H = 1, O = 16, Na = 23, S = 32, Cu = 64]

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Al4C3 + 12H2O ⟶ 4Al(OH)3 + 3CH4
(i) State what mass of aluminium hydroxide is formed from 12 g. of aluminium carbide.
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(ii) the quantity in moles of N₂ formed.
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