# What is [HCl(aq)] in the following scenario?

## A $1.12 \cdot L$ volume of carbon dioxide was collected under standard conditions from the reaction between $H C l \left(a q\right)$ and EXCESS $\text{calcium carbonate}$?

Mar 26, 2016

We assume conditions of $1$ $a t m$, and a temperature of $298 K$. The normality of $H C l$ is approx. $1 \cdot m o l \cdot {L}^{-} 1$.

#### Explanation:

We always start these sorts of problems with a balanced equation:

$C a C {O}_{3} \left(s\right) + 2 H C l \left(a q\right) \rightarrow C a C {l}_{2} \left(a q\right) + {H}_{2} O \left(l\right) + C {O}_{2} \left(g\right) \uparrow$

From the stoichiometric equation, it is clear that whatever quantity of carbon dioxide gas is collected, this is HALF the molar quantity of acid added.

Now, the molar volume of an ideal gas at $298 K$ is $24.5 \cdot L$.

So $\text{moles of "CO_2" collected}$ $=$ $\frac{1.12 \cdot \cancel{L}}{24.5 \cdot \cancel{L} \cdot m o {l}^{-} 1}$ $=$ $0.0457 \cdot m o l$

To generate this quantity of gas, there MUST have been TWICE this molar quantity of $H C l$ in the $100$ $m L$ volume initially added to the carbonate.

So $\text{Concentration of acid}$ $=$ $\frac{0.0457 \cdot m o l \times 2}{100 \times {10}^{-} 3 L}$ $=$ ??mol*L^-1. A more accurate answer would be obtained if we included $\text{SVP}$, the saturated vapour pressure of water under the given conditions (i.e. a gas collected under water, will always include the $\text{SVP}$).