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

Question

Ellie Andrews · Accepted 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.

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.) https://tutor.hix.ai In order to answer this question, divide the given volume by the molar volume of 27.710980 L/mol. https://tutor.hix.ai #"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://tutor.hix.ai 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)

Cooper Anderson · Answer

undefined 2.85 moles.