Question

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

Answer 1

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)

Answer 2

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"`