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 \(0.14\times 2=0.28\) 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 \(CO_2\).
\(\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2\)
\(\text{Mass of }CO_2=(0.14mole)\times (44g/mole)=6.16g\)
Therefore, the mass of \(CO_2\) produced will be, 6.16 grams.
