# Calculate the volume of carbon dioxide gas lost at 25 degrees and 100kPa?

## Initial mass (g) = 125.5 Final mass (g) = 124.8 Change in mass (g) = 0.7

Feb 3, 2018

$0.38 L$, or $380 c {m}^{3}$ lost.

#### Explanation:

Use the Ideal Gas Law:

$P V = n R T$

For the initial mass of $C {O}_{2}$:

Remember, the formula for moles, $n$, is $n = \frac{m}{M}$, where $m$ is the given mass and $M$ the molar mass.

$M = 44.01 \frac{g}{\text{mol}}$

Here, $m = 125.5 g$.

So $n = \frac{125.5}{44.01} = 2.85 \text{mol}$.

So now, we can rearrange the gas law:

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

$V = \frac{2.85 \cdot 8.314 \cdot 298.15}{100}$

$V = 70.65 L$ of carbon dioxide before the loss.

For the final mass of $C {O}_{2}$:

Here, $m = 124.8 g$

So $n = \frac{124.8}{44.01} = 2.835 \text{mol}$

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

$V = \frac{2.835 \cdot 8.314 \cdot 298.15}{100}$

$V = 70.27 L$ of carbon dioxide after loss.

Therefore, the total loss must be ${V}_{\text{final"-V_"initial}}$.

So the loss is equal to $70.65 - 70.27 = 0.38 L$.