Answer : The mass of CO_2 produced will be, 6.16 grams.

Explanation : Given,

Mass of CaCO_3 = 14 g

Mass of HCl = 56.70 g

Molar mass of CaCO_3 = 100 g/mole

Molar mass of HCl = 36.5 g/mole

Molar mass of CO_2 = 44 g/mole

First we have to calculate the moles of CaCO_3 and HCl.

text{Moles of }CaCO_3=frac{text{Mass of }CaCO_3}{text{Molar mass of }CaCO_3}=frac{14g}{100g/mole}=0.14moles

text{Moles of }HCl=frac{text{Mass of }HCl}{text{Molar mass of }HCl}=frac{56.70g}{36.5g/mole}=1.55moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCO_3(s)+2HCl(aq)rightarrow CO_2(g)+H_2O(l)+CaCl_2(aq)

From the balanced reaction we conclude that

As, 1 moles of CaCO_3 react with 2 mole of HCl

So, 0.14 moles of CaCO_3 react with 0.14times 2=0.28 moles of HCl

From this we conclude that, HCl is an excess reagent because the given moles are greater than the required moles and CaCO_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of CO_2.

As, 1 moles of CaCO_3 react to give 1 moles of CO_2

So, 0.14 moles of CaCO_3 react to give 0.14 moles of CO_2

Now we have to calculate the mass of CO_2.

text{Mass of }CO_2=text{Moles of }CO_2times text{Molar mass of }CO_2

text{Mass of }CO_2=(0.14mole)times (44g/mole)=6.16g

Therefore, the mass of CO_2 produced will be, 6.16 grams.