What is the volume of 5.0 grams of #CO_2# at STP?

2 Answers
May 18, 2017

Answer:

The volume of #"5.0 g CO"_2"# is #"2.6 L CO"_2"# at STP.

Explanation:

#color(blue)("STP"#

STP is currently #0^@"C"# or #"273.15 K"#, which are equal, though the Kelvin temperature scale is used for gas laws; and pressure is #"10"^5color(white)(.)"Pascals (Pa)"#, but most people use #"100 kPa"#, which is equal to #10^5color(white)(.)"Pa"#.

You will use the ideal gas law to answer this question. Its formula is:

#PV=nRT#,

where #P# is pressure, #V# is volume, #n# is moles, #R# is a gas constant, and #T# is temperature in Kelvins.

#color(blue)("Determine moles"#
You may have noticed that the equation requires moles #(n)#, but you have been given the mass of #"CO"_2"#. To determine moles, you multiply the given mass by the inverse of the molar mass of #"CO"_2"#, which is #"44.009 g/mol"#.

#5.0color(red)cancel(color(black)("g CO"_2))xx(1"mol CO"_2)/(44.009color(red)cancel(color(black)("g CO"_2)))="0.1136 mol CO"_2"#

#color(blue)("Organize your data"#.

Given/Known
#P="100 kPa"#
#n="0.1136 mol"#
#R="8.3145 L kPa K"^(-1) "mol"^(-1)"#
https://en.wikipedia.org/wiki/Gas_constant
#T="273.15 K"#

Unknown: #V#

#color(blue)("Solve for volume using the ideal gas law."#
Rearrange the formula to isolate #V#. Insert your data into the equation and solve.

#V=(nRT)/P#

#V=(0.1136color(red)cancel(color(black)("mol"))xx8.3145 color(white)(.)"L" color(red)cancel(color(black)("kPa")) color(red)cancel(color(black)("K"))^(-1) color(red)cancel(color(black)("mol"))^(-1)xx273.15color(red)cancel(color(black)("K")))/(100color(red)cancel(color(black)("kPa")))="2.6 L CO"_2"# rounded to two significant figures due to #"5.0 g"#

May 18, 2017

Answer:

I got 2.55 Liters

Explanation:

1 mole of any gas at STP = 22.4 Liters
#5 g CO_2(g) = (5 g)/(44(g/"mole")) = 0.114 "mole" CO_2(g) #
Volume of 0.114 mole #CO_2(g)# = (0.114 mole)(22.4 L/mole) = 2.55 Liters #CO_2#(g) at STP