Question

Calcium carbonate reacts with aqueous $ {\text{HCl}} $ to give $ {\text{CaC}}{{\text{l}}_{\text{2}}} $ and $ {\text{C}}{{\text{O}}_{\text{2}}} $ according to the reaction,
 $ {\text{CaC}}{{\text{O}}_{\text{3}}}{\text{(s) + 2HCl(aq) }} \to {\text{ CaC}}{{\text{l}}_{\text{2}}}{\text{(aq) + C}}{{\text{O}}_{\text{2}}}{\text{(g) + }}{{\text{H}}_{\text{2}}}{\text{O(l)}} $
What mass of $ {\text{CaC}}{{\text{O}}_{\text{3}}} $ is required to react completely with $ {\text{25mL}} $ of $ {\text{0}}{\text{.75M HCl}} $ ?

Answer 1

Hint: In the above question, since we have to find the mass of $ {\text{CaC}}{{\text{O}}_{\text{3}}} $ , so first we have to find out number of moles of $ {\text{CaC}}{{\text{O}}_{\text{3}}} $ present . For finding number of moles of $ {\text{CaC}}{{\text{O}}_{\text{3}}} $ we have to find out number of moles of $ {\text{HCl}} $ which can be found out by molarity formula.
Formula Used
Molarity= $ \dfrac{{\text{n}}}{{\text{V}}} $
Where n is the number of moles of the solute and V is the volume of solution given in litre.

Complete step by step solution:
In the above question, a chemical equation is given:
 $ {\text{CaC}}{{\text{O}}_{\text{3}}}{\text{(s) + 2HCl(aq) }} \to {\text{ CaC}}{{\text{l}}_{\text{2}}}{\text{(aq) + C}}{{\text{O}}_{\text{2}}}{\text{(g) + }}{{\text{H}}_{\text{2}}}{\text{O(l)}} $
So, our first step is to check whether the equation is balanced or not. Since, the equation is balanced, we find that 1 mole of $ {\text{CaC}}{{\text{O}}_{\text{3}}} $ reacts with 2 mole of $ {\text{HCl}} $ .
So, now we have to find out how many moles of $ {\text{HCl}} $ is present in the solution.
We know that:
Molarity= $ \dfrac{{\text{n}}}{{\text{V}}} $
By cross-multiplication, we get:
n= Molarity $ \times $ V
Substituting the molarity and volume of $ {\text{HCl}} $ solution, we get:
n= $ 0.75 \times \dfrac{{25}}{{1000}}{\text{ = 0}}{\text{.01875}} $
So, the number of moles of $ {\text{HCl}} $ is present is $ {\text{0}}{\text{.01875}} $ .
Since , 2 moles of $ {\text{HCl}} $ reacts with 1 mole of $ {\text{CaC}}{{\text{O}}_{\text{3}}} $ .
So, $ {\text{0}}{\text{.01875}} $ moles of $ {\text{HCl}} $ reacts with $ {\text{0}}{{.01875 \times }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = 0}}{\text{.009375}} $ moles of $ {\text{CaC}}{{\text{O}}_{\text{3}}} $ .
Molar mass of $ {\text{CaC}}{{\text{O}}_{\text{3}}} $ = atomic mass of Ca + atomic mass of C + 3 $ \times $ atomic mass of O
So, Molar mass of $ {\text{CaC}}{{\text{O}}_{\text{3}}} $ = $ 40 + 12 + 3 \times 16 = 52 + 48 = 100 $
Mass of $ {\text{CaC}}{{\text{O}}_{\text{3}}} $ present in 1 mole = 100g
So, Mass of $ {\text{CaC}}{{\text{O}}_{\text{3}}} $ present in $ {\text{0}}{\text{.009375}} $ moles = $ {\text{0}}{{.009375 \times 100 = 0}}{\text{.9375g}} $ .
 $ \therefore $ Mass of $ {\text{CaC}}{{\text{O}}_{\text{3}}} $ is required to react completely with $ {\text{25mL}} $ of $ {\text{0}}{\text{.75M HCl}} $ is $ {\text{0}}{\text{.9375g}} $ .

Note:
While solving these types of questions, the first step should be to check whether the given equation is balanced or not.Weight of the substance present in its 1 mole is equal to the molar mass of that substance.As molarity is dependent on volume, it is also dependent on the temperature.