STP: What volume of gas (in L) is occupied by 16.0g of CH4 at STP?
Is the answer: 22.3L? I need to validate that I've completed the problem correctly. I'm not sure if I am using the appropriate formula! THANKS!
- Old Science GuyLv 74 years agoFavorite Answer
the molar volume of a gas at STP is 22.4 L/mol
so convert mass to moles then convert moles to volume at STP
16.0 g CH4 * (1 mol / 16.0 g CH4) * (22.4 L/mol) = 22.4 L
not sure where you got 22.3
maybe you rounded incorrectly
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- hcbiochemLv 74 years ago
To demonstrate WHY 1 mole of any gas has a volume of 22.4 L at STP, use the ideal gas law:
PV = n RT
(1 atm) V = 1 mol (0.0821 Latm/molK) (273 K)
V = 22.4 L
In your question, as others have shown, 16.0 g CH4 = 1 mol CH4.
- 4 years ago
1 mole of any gas occupies 22.4 L in STP,
V=(16/M) * 22.4
M is the molecular weight
- Fred KLv 74 years ago
1 mole occupies 22.4 L at stp
as one decimal place is given then 16.0g (16.04g) is still one mole.