# How many moles of CO_2 (g) are in a 5.6 L sample of CO_2 measured at STP?

Jan 30, 2016

$\text{moles} = 0.25$

#### Explanation:

Using the ideal gas equation, we can solve for the number of moles:

$P V = n R T$

where:
$P =$pressure
$V =$volume
$n =$moles
$R =$universal constant $\left(8.314 \frac{k P a \cdot L}{m o l \cdot K}\right)$
$T =$temperature (Kelvin)

Recall that at STP conditions:

$P = 101.325$ $k P a$
$T = 273.15$ $K$

To solve for the number of moles of carbon dioxide gas, substitute your known values into the ideal gas equation:

$P V = n R T$

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

$n = \frac{\left(101.325 k P a\right) \left(5.6 L\right)}{\left(8.314 \frac{k P a \cdot L}{m o l \cdot K}\right) \left(273.15 K\right)}$

$n = \frac{\left(101.325 \textcolor{red}{\cancel{\textcolor{b l a c k}{k P a}}}\right) \left(5.6 \textcolor{\mathmr{and} a n \ge}{\cancel{\textcolor{b l a c k}{L}}}\right)}{\left(8.314 \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{k P a}}} \cdot \textcolor{\mathmr{and} a n \ge}{\cancel{\textcolor{b l a c k}{L}}}}{m o l \cdot \textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{K}}}}\right) \left(273.15 \textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{K}}}\right)}$

$n = 0.2498580892$ $m o l$

$n = 0.25$ $m o l$ (rounded to $2$ significant figures)

$\therefore$, there are $0.25$ $m o l$ in $5.6$ $L$ of $C {O}_{2 \left(g\right)}$ at STP conditions.