How many moles of gas are in a volume of 63.3 L at STP?

1 Answer
Jul 31, 2016

Answer:

At STP of 273.15 K and 100 kPa, 63.3 L of an ideal gas would contain 2.33 moles.

At STP of 273.15 K and 1 atm, 63.3 L of an ideal gas would contain 2.82 moles.

Explanation:

At STP of 273.15 K #("0"^@"C")# and a pressure of #10^5# Pa, usually written as 100 kPa, the molar volume of an ideal gas is 22.710980 L/mol. (This STP has been recommended by the International Union of Pure and Applied Chemistry (IUPAC) since 1990.) http://goldbook.iupac.org/S05910.html

In order to answer this question, divide the given volume by the molar volume of 27.710980 L/mol.
https://en.m.wikipedia.org/wiki/Molar_volume

#"63.3 L"/"27.710980 mol/L"="2.33 moles"# (rounded to three significant figures)

Just in case your teacher uses the older values for STP, 273.15 K and 1 atm. The molar volume of a gas at this STP is 22.414 mol/L.
https://en.m.wikipedia.org/wiki/Molar_volume

Just as with the previous calculation, divide the given volume by the molar volume of 22.414 mol/L.

#"63.3 L"/"22.414 mol/L"="2.82 moles"# (rounded to three significant figures)