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

Answer 1

To calculate the volume occupied by 55 g of methane (CH4) at STP (Standard Temperature and Pressure), we first need to determine the number of moles of methane. The molar mass of methane is 16 g/mol. Therefore, 55 g of methane is equal to 55 g / 16 g/mol ≈ 3.44 mol. At STP, 1 mol of any gas occupies approximately 22.4 L. Therefore, 3.44 mol of methane would occupy approximately 3.44 mol x 22.4 L/mol ≈ 77.1 L.

Answer 2

Oh, dude, at STP (Standard Temperature and Pressure), one mole of any gas occupies 22.4 liters. So, to find the volume of 55 grams of methane (CH4), we first need to convert grams to moles using the molar mass of methane. Then, we can use the ideal gas law to calculate the volume. It's like a math problem wrapped in a science joke!

Answer 3

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

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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

Answer 4

At STP (standard temperature and pressure), the volume occupied by 1 mole of any ideal gas is 22.4 L. The molar mass of methane (CH₄) is 16 g/mol. Therefore, 55 g of methane is approximately 3.44 moles. Thus, the volume occupied by 55 g of methane at STP is around 78.4 L.

Answer 5

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.

Answer 6

The volume of methane is 83,84 L.

Answer 7

130L