# Given a 20*g mass of calcium carbonate, how much carbon dioxide will be generated, when it is added to a 100*mL volume of "20% w/w" solution of HCl(aq)?

Aug 19, 2017

The units of concentration of the hydrochloric acid are unfortunate.....I gets approx. a $5 \cdot L$ volume under standard conditions.

#### Explanation:

We interrogate the reaction......

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

Now this site reports that the density of 20% $H C l$ is $\rho = 1.098 \cdot g \cdot m {L}^{-} 1$. These data SHOULD have been supplied with the question.

And thus moles of HCl=(20%xx100*cancel(mL)xx1.098*cancelg*cancel(mL^-1))/(36.46*cancelg*mol^-1)

$= 0.602 \cdot \frac{1}{m o {l}^{-} 1} = 0.602 \cdot \frac{1}{\frac{1}{m o l}} = 0.602 \cdot m o l$

And thus the molar concentration of the hydrochloric acid was approx. $6 \cdot m o l \cdot {L}^{-} 1$.

And also moles of $\text{calcium carbonate} \equiv \frac{20 \cdot g}{100.09 \cdot g \cdot m o {l}^{-} 1}$

$= 0.200 \cdot m o l$.

There is thus a stoichiometric excess of hydrochloric acid, and calcium carbonate is the limiting reagent. A $0.200 \cdot m o l$ quantity of carbon dioxide gas will be released.

Depending on your syllabus, we know that the molar volume at $1 \cdot a t m$ and $298 \cdot K$ is $24.5 \cdot L \cdot m o {l}^{-} 1$. You have to check with the given definitions in your course.

And thus "volume"=24.5*L*mol^-1xx0.200*mol=??*L.

How much acid is in excess?