What is the volume of 4.5 g of #"CO_2"#?

2 Answers
Dec 1, 2015

Answer:

The volume of #"4.5 g CO"_2"# is #"2300 cm"^3"# at STP.

Explanation:

At STP, #"273.15 K"# and #"1 atm"#, the density of carbon dioxide is #"0.001977 g/cm"^3"#.
http://www.engineeringtoolbox.com/gas-density-d_158.html

We can use the given mass and known density to determine the volume of #"CO"_2"# at STP. Divide the mass by the density.

#4.5cancel"g CO"_2xx(1"cm"^3 "CO"_2)/(0.001977cancel"g CO"_2)="2300 cm"^3 "CO"_2"# (rounded to two significant figures)

Dec 1, 2015

Answer:

The volume of #"4.5 g CO"_2"# is #"2300 cm"^3"# at STP.

Explanation:

We can also use the ideal gas law to solve this problem.
The equation is #PV=nRT#, where #n# is the number of moles, and #R# is the gas constant.

In this problem, #T# and #P# are at STP, #"273.15 K"# and #"1 atm"#.

Determine moles of #"CO"_2"# by dividing the given mass by the molar mass.

#4.5cancel"g CO"_2xx(1"mol CO"_2)/(44.01cancel"g CO"_2)="0.10225 mol CO"_2"#

I am leaving some guard digits to reduce rounding errors.

Ideal Gas Law

Given/Known
#P="1 atm"#
#n="0.10225 mol"#
#R="0.082057338 L atm K"^(-1) "mol"^(-1)"#
https://en.m.wikipedia.org/wiki/Gas_constant
#T="273.15 K"#

Unknown
Volume, #V#

Equation
#P_1V_1=nRT#

Solution
Rearrange the equation to isolate #V# and solve.

#V=(nRT)/(P)#

#V=((0.10225cancel"mol"xx0.082057338 "L" cancel"atm" cancel("K"^(-1)) cancel("mol"^(-1))xx273.15cancel"K"))/(1cancel"atm")="2.3 L"# (rounded to two significant figures)

Convert to #"cm"^3"#.

#2.3cancel"L"xx(1000cancel"mL")/(1cancel"L")xx(1"cm"^3)/(1cancel"mL")="2300 cm"^3"#