What volume of oxygen is needed to burn 8g of
magnesium at r.t.p
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2
Answer:
Given volume of methane =0.25dm
3
=0.25L
Combustion of methane (CH
4
)-
(methane)
CH
4
+2O
2
⟶CO
2
+2H
2
O
As we know that volume of 1 mole of gas is 22.4 L.
Amount of oxygen required for the combustion of 22.4 L of methane =(2×22.4)L=44.8L
∴ Amount of oxygen required for the combustion of 0.25 L of methane =
22.4
44.8
×0.25=0.5L
Hence, 0.5 L of oxygen required for the complete combustion of 0.25dm
3
of methane.
Explanation:
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