# Calcium carbonate and hydrochloric acid

I am trying to solve an exercise, where a block of 605.5g calcium carbonate should be completely dissolved in 30% hydrochloric acid (w/w) and the concentration of the acid should be 3% in the end. The question is: How many g of 30% HCl-acid is needed in the beginning.

$$\ce{CaCO3 + 2 HCl -> CaCl2 + H2O + CO2}$$

So for each dissolved molecule of $\ce{CaCO3}$:

• 2 molecules of $\ce{HCl}$ are used
• 1 molecule of $\ce{H2O}$ is produced
• 1 molecule of $\ce{CaCl2}$ is produced
• 1 molecule of $\ce{CO2}$ is produced, but leaves the solution

I got a result (about $\pu{1671.524g}$ 30% hydrochloric acid), which I verified this way:

Let $p$ be the number of dissolved $\ce{CaCO3}$ moles, which is about $\mathrm{605.5/100.0869 = 6.05}$

$\ce{HCl = 1671.524g * 0.3}$

$\ce{H_2O = 1671.524g * 0.7}$

$\ce{HCl_{end} = HCl - 2 * (p * M(HCl)}$)

$\ce{H2O_{end} = H2O + p * M(H2O)}$

$\ce{CaCl_{2end} = p * M(CaCl2)}$

$c_{end}$ = $\ce{\frac{HCl_{end}}{HCl_{end} + H2O_{end} + CaCl_{2end}} = 0.03}$

But my solution is claimed to be incorrect. Am I missing something here?

• You need to tell us what % means in your question. Is it $\mathrm (w/w)$ or $\mathrm (w/v)$ since the solvent is $\ce {H2O}$? That make a difference as well. – Mathew Mahindaratne Apr 2 '18 at 18:13
• @MathewMahindaratne it is w/w, I added that in the question – Ctx Apr 2 '18 at 18:21
• Okay, now what is your objective? Is it to find how many $\mathrm {g}$ of $\ce {HCl}$ you have to use? I think that is the most obvious. – Mathew Mahindaratne Apr 2 '18 at 18:35
• @MathewMahindaratne Yes, the g of hcl – Ctx Apr 2 '18 at 18:54
• Your approach is i n correct track, but your calculations probably need attention. I'll post an answer in few minutes. – Mathew Mahindaratne Apr 2 '18 at 19:18

We'll assume the reaction loses only $\ce {CO2}$ from the system (although it is a exothermic reaction, we'll assume $\ce{H2O}$ produced would not be lost as vapors): $$\ce{CaCO3 + 2 HCl -> CaCl2 + H2O + CO2}$$

Mass of $30\%~\ce{HCl}$ needed to react with $\pu {605.5 g}$ of $\ce{CaCO3}$ $\pu {= 605.5 g \ce{CaCO3} \times \frac{\pu{1mol}~\ce{CaCO3}}{\pu{100.09g}~ \ce{CaCO3}}\times \frac{\pu{2mol}~\ce{HCl}}{\pu{1 mol}~\ce{CaCO3}}\times \frac{\pu{36.45g}~\ce{HCl}}{\pu{1mol}~\ce{HCl}}\times \frac{\pu{100g}~\text{solution of 30%}~\ce{HCl}}{\pu{30 g}~\ce{HCl}}= 1470.04 g of 30\% \ce{HCl} solution}$.

Similarly, mass of $\ce{CO2}$ released when $\ce{HCl}$ reacted with $\pu{605.5 g}$ of $\ce{CaCO3}$ $\pu{= 605.5 g \ce{CaCO3} \times \frac{\pu{1mol}~\ce{CaCO3}}{\pu{100.09g}~\ce{CaCO3}}\times \frac{\pu{1mol}~\ce{CO2}}{\pu{1mol}~\ce{CaCO3}}\times \frac{\pu{44.0g}~\ce{CO2}}{\pu{1mol}~\ce{CO2}} = 266.18 g \ce{CO2}}$.

According to the law of conservation of mass,

$\pu {mass of reactant = mass of products}$

Assuming no solvent left the system,

$\pu {mass of reactant + solvent = mass of products + solvent}$

$\pu{mass of reactant + solvent = 1470.04 g + 605.5 g = 2075.54 g}$

$\pu{mass of products + solvent remained in the flask = mass of reactant + solvent - mass of \ce{CO2} released = 2075.54 g - 266.18 g = 1809.41 g}$

Now, suppose extra $\pu{A g}$ of $30\%~\ce{HCl}$ added to the solution, so that final concentration is $3\%~\ce {HCl}$.

$\pu {mass of \ce{HCl} in A g of 30\% solution = A g of 30\% \ce{HCl} solution \times \frac {\pu{30g}~\ce {HCl}}{\pu{100g}~ \text{of 30%$\ce{HCl}$solution}} = 0.3A g of 30\% \ce{HCl}}$

Thus, since final concentration of solution is $\mathrm {3\%}$,

$$\pu {\frac {(0.3 A) g}{(1809.41 + A) g} = 0.03}$$ $$\pu {0.3 A = 0.03 \times 1809.41 + 0.03A}$$ Thus, $\mathrm {A} = \frac {0.03 \times 1809.41}{0.3-0.03} = \pu{201.04 g}$

Finally, the mass of $30\%~\ce{HCl}$ initially added is $\pu {(1470.04 + 201.04) g = 1671.1 g}$.

• Hello, @Mathew Mahindaratne - can you elaborate on what does "your reaction loses only CO2 from the system" means? That is gasifys and exits the solution? – Stipe Galić Aug 17 '18 at 14:45
• @Stipe Galić: Yes, therefore, mass of $\ce{CO2}$ should be eliminated from calculations. – Mathew Mahindaratne Aug 17 '18 at 16:12