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

Dec 1, 2015

The volume of $\text{4.5 g CO"_2}$ is $\text{2300 cm"^3}$ at STP.

#### Explanation:

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

We can use the given mass and known density to determine the volume of $\text{CO"_2}$ at STP. Divide the mass by the density.

$4.5 \cancel{\text{g CO"_2xx(1"cm"^3 "CO"_2)/(0.001977cancel"g CO"_2)="2300 cm"^3 "CO"_2}}$ (rounded to two significant figures)

Dec 1, 2015

The volume of $\text{4.5 g CO"_2}$ is $\text{2300 cm"^3}$ at STP.

#### Explanation:

We can also use the ideal gas law to solve this problem.
The equation is $P V = n R T$, where $n$ is the number of moles, and $R$ is the gas constant.

In this problem, $T$ and $P$ are at STP, $\text{273.15 K}$ and $\text{1 atm}$.

Determine moles of $\text{CO"_2}$ by dividing the given mass by the molar mass.

$4.5 \cancel{\text{g CO"_2xx(1"mol CO"_2)/(44.01cancel"g CO"_2)="0.10225 mol CO"_2}}$

I am leaving some guard digits to reduce rounding errors.

Ideal Gas Law

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

Unknown
Volume, $V$

Equation
${P}_{1} {V}_{1} = n R T$

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

$V = \frac{n R T}{P}$

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 $\text{cm"^3}$.

$2.3 \cancel{\text{L"xx(1000cancel"mL")/(1cancel"L")xx(1"cm"^3)/(1cancel"mL")="2300 cm"^3}}$