What is the mass of "5.6 L"5.6 L of "O"_2"O2 at 23^@"C"23C and "8.16 atm"8.16 atm?

1 Answer
Jun 1, 2017

The mass of "O"_2O2 is "60.2 g"60.2 g

Explanation:

This can be solved by using the Ideal Gas Law:

PV=nRTPV=nRT,

where PP is pressure, VV is volume, nn is moles, RR is the universal gas constant, and TT, which is the temperature in Kelvins.

First calculate the number of moles of "O"_2"O2, using the equation for the Ideal Gas Law. Then multiply the moles "O"_2O2 by its molar mass to get the mass of "O"_2"O2.

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

Known/Given

P="8.16 atm"P=8.16 atm

V="5.60 L"V=5.60 L

R="0.08206 L atm K"^(-1) "mol"^(-1)R=0.08206 L atm K1mol1

T="23"^@"C"+273="296 K"T=23C+273=296 K

color(blue)("Moles of Oxygen Gas"Moles of Oxygen Gas

Rearrange the equation to isolate nn. Insert your data and solve.

n=(PV)/(RT)n=PVRT

n=(8.16color(red)cancel(color(black)("atm"))xx5.60color(red)cancel(color(black)("L")))/(0.08206 color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) color(red)cancel(color(black)("K"))^(-1) "mol"^(-1)xx296color(red)cancel(color(black)("K")))="1.88 mol"

There are "1.88 mol O"_2" under the conditions described in the question.

color(blue)("Mass of Oxygen Gas"

Multiply the mol "O"_2" by its molar mass, "31.998 g/mol".

1.88color(red)cancel(color(black)("mol O"_2))xx(31.998"g O"_2)/(1color(red)cancel(color(black)("mol O"_2)))="60.2 g O"_2 rounded to three sig figs