Chemistry, asked by Chris9687, 7 months ago

Calculate the volume of oxygen required at STP for complete combustion of one mole of methane

Answers

Answered by karthikeyan0175
0

Answer:

The volume of oxygen required is

0.50 dm

3

.

Explanation:

For this problem, we can use Gay-Lussac's Law of Combining Volumes:

If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.

The balanced equation for the combustion is

l

CH

4

+

2O

2

CO

2

+

2H

2

O

1 dm

3

l

l

2 dm

3

According to Gay-Lussac,

1 dm

3

l

of CH

4

requires

2 dm

3

l

of O

2

.

Volume of O

2

=

0.25

dm

3

l

CH

4

×

2 dm

3

l

O

2

1

dm

3

l

CH

4

=

0.50 dm

3

l

O

2

Answered by anveshadeshmukh68
1

CH4 +2O2 ⟶CO2 +2H2O

1 mole CH 4 requires 2×22.4 L O 2 at STP.

0.25 moles CH 4 =0.25×2×22.4

=11.2 L or 11.2 dm 3

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