# Question #7baa6

Jun 21, 2017

Well, we know that $1 \cdot m o l$ of Ideal gas at $\text{STP}$ occupies $22.4 \cdot L$...

#### Explanation:

And thus if we assume that carbon dioxide behaves ideally (which is a reasonable assumption, then we have $\frac{224 \cdot L}{22.4 \cdot L \cdot m o {l}^{-} 1}$, i.e. a $10 \cdot m o l$ quantity of $C {O}_{2}$.....And I think you can see why they specified this particular volume.

And thus there is a $10 \cdot m o l$ quantity of carbon, i.e. $10 \cdot m o l \times 12.00 \cdot g \cdot m o {l}^{-} 1 = 120.0 \cdot g$........

And likewise there is $20 \cdot m o l \times 15.999 \cdot g \cdot m o {l}^{-} 1 = 320.0 \cdot g$........with respect to $\text{oxygen}$.

And you should not have to remember the atomic masses (tho you will soon retain the common masses) because you should be provided with a Periodic Table.