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Calculate the minimum volume of oxygen gas, at room temperature and pressure(r.t.p). required to completely burn 56g of methane gas. Give your answer to two significant figures. (1 mole of any gas occupies 24 dm³ at r.t.p,Ar:C=12,H=1)


BALANCED EQUATION FOR THE BURNING OF METHANE GAS.

CH₄+2O₂→CO₂+2H₂O

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Answer:

170 dm³ (2 significant figures)

Explanation:

CH₄ + 2O₂ → CO₂ + 2H₂O

We need to find how much 'volume' of O₂ is required to burn 56 g of CH₄.

Let's take the ratios of the two gases only.

CH₄ : 2O₂ (in terms of moles the ratio is 1:2)

56g : x dm³

Mr of CH₄ : Volume of oxygen at rtp

Find the Mr of CH₄.

Mr of CH₄ = 12 + (1 × 4) = 16

We know that 1 mole of any gas occupies 24 dm³ at room temperature and pressure (rtp). We have to find the volume of oxygen, there are 2 moles of oxygen, so 2 moles of gas will occupy:

2 × 24 = 48 dm³

56g : x dm³

  16 : 48 dm³

16x = 56 × 48

16x = 2688

x = [tex]\frac{2688}{16}[/tex]

x = 168 dm³

∴ the minimum volume of oxygen gas, at room temperature and pressure required to completely burn 56g of methane gas is 170 dm³

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