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The easiest way to approach this is to remember molar volume: one mole of every (ideal) gas will occupy 22.4 liters at STP - a gas occupies 22.4 L/mol.

Since we know that, we can set up a simple proportion that follows the thought process: if one mole of a gas occupies 22.4 liters (at STP), then 2.88 grams of methane occupies x liters.

But before we do that, we have a problem: the measurement given is in the wrong units - it is in grams when molar volume is in moles. Therefore, we first need to convert 2.88 grams of methane to moles. To do this, we need the molar mass of the compound - the sum of all the atomic masses involved.

Carbon = 12.0 grams

Hydrogen = 1.01 grams × 4 atoms = 4.04 grams

-------------------------------------------------------------

Methane = 16.04 grams

To convert grams to moles:

Grams of substance ÷ Molar mass (in grams) = Moles of substance

2.88 grams CH4 ÷ 16.04 grams CH4 = 0.180 moles CH4

Now we follow through with our proportion:

22.4 L/1 mol = x L/.180 mol

x = 4.03 L

2.88 grams of methane occupies 4.03 liters at STP

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14y ago
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13y ago

38.0 g F2 = 1.00 mole F2 (divide 38.0 by molar mass= 2*19.00 g/mol)

At STP each (1.00) mole of ANY gas will occupy 22,4 Litre.

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8y ago

The volume of methane is 83,84 L.

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11y ago

130L

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Q: What volume is occupied by 55 g of methane CH4 (g) at STP?
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