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