Question

How much volume of dioxygen will be required for complete combustion 0.08g of methane​
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Answer 1

ᗩᑎՏᗯᗴᖇ✈

1 mole of a gas occupies 22.4 L of a gas at STP.

1 mole of a gas occupies 22.4 L of a gas at STP.Since, the molecular weight of methane is 16 g,

1 mole of a gas occupies 22.4 L of a gas at STP.Since, the molecular weight of methane is 16 g, Therefore, at STP, 22.4 L corresponds to 16 g.

1 mole of a gas occupies 22.4 L of a gas at STP.Since, the molecular weight of methane is 16 g, Therefore, at STP, 22.4 L corresponds to 16 g.112 cm

1 mole of a gas occupies 22.4 L of a gas at STP.Since, the molecular weight of methane is 16 g, Therefore, at STP, 22.4 L corresponds to 16 g.112 cm 3

1 mole of a gas occupies 22.4 L of a gas at STP.Since, the molecular weight of methane is 16 g, Therefore, at STP, 22.4 L corresponds to 16 g.112 cm 3 or 0.112 L will correspond to

1 mole of a gas occupies 22.4 L of a gas at STP.Since, the molecular weight of methane is 16 g, Therefore, at STP, 22.4 L corresponds to 16 g.112 cm 3 or 0.112 L will correspond to 22.4

1 mole of a gas occupies 22.4 L of a gas at STP.Since, the molecular weight of methane is 16 g, Therefore, at STP, 22.4 L corresponds to 16 g.112 cm 3 or 0.112 L will correspond to 22.416

1 mole of a gas occupies 22.4 L of a gas at STP.Since, the molecular weight of methane is 16 g, Therefore, at STP, 22.4 L corresponds to 16 g.112 cm 3 or 0.112 L will correspond to 22.416

1 mole of a gas occupies 22.4 L of a gas at STP.Since, the molecular weight of methane is 16 g, Therefore, at STP, 22.4 L corresponds to 16 g.112 cm 3 or 0.112 L will correspond to 22.416 ×0.112=0.08g.

Answer 2

Answer:

1 mole of a gas occupies 22.4 L of a gas at STP.

Since, the molecular weight of methane is 16 g,

Therefore, at STP, 22.4 L corresponds to 16 g.

112 cm

3

or 0.112 L will correspond to

22.4

16

×0.112=0.08g.