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How many grams of carbon are required to form 44.8 dm³ of methane gas at STP​

User EAK TEAM
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1 Answer

3 votes

Answer:

24 g of carbon

Step-by-step explanation:

Methane is produced by this reaction:

C(s) + 2H₂(g) → CH₄ (g)

At STP 1 mol of any gas is contained at 22.4L. Let's see how many moles are contained at 44.8L

(1L = 1dm³)

44.8 L . 1mol / 22.4L = 2 moles

Stoichiometry at reaction is 1:1.

1 mol of methane is produced by 1 mol of carbon.

In conclussion, 2 moles of gas may come from 2 moles of C.

We determine the mass:

2 mol . 12 g /mol = 24 g

User Alexandre Juma
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