Calcium carbonate and hydrochloric acid
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): 
$$\ce{CaCO3 + 2 HCl -&gt; CaCl2 + H2O + CO2}$$

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 $\pu {(1470.04 + 201.04) g = 1671.1 g}$.