How do you calculate the volume of 25.0 g of carbon monoxide at STP?

1 Answer
Jul 16, 2016

Answer:

The volume is 20.3 L.

Explanation:

We can convert the mass of carbon monoxide to moles and then use the Ideal Gas Law to calculate the volume at STP.

Moles of #"CO"#

#"Moles of CO" = 25.0 color(red)(cancel(color(black)("g CO"))) × ("1 mol CO")/(28.01 color(red)(cancel(color(black)("g CO")))) = "0.8925 mol CO"#

Volume at STP

The Ideal Gas Law is:

#color(blue)(|bar(ul(PV = nRT)|)#,

where

  • #P# is the pressure
  • #V# is the volume
  • #n# is the number of moles
  • #R# is the gas constant
  • #T# is the temperature

We can rearrange the Ideal Gas Law to get

#V = (nRT)/P#

STP is 1 bar and 0 °C.

#n = "0.8925 mol"#
#R = "0.083 14 bar·L·K"^"-1""mol"^"-1"#
#T = "273.15 K"#
#P = "1 bar"#

#V = (nRT)/P = (0.8925 color(red)(cancel(color(black)("mol"))) × "0.083 14" color(red)(cancel(color(black)("bar")))·"L"· color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = "20.3 L"#