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

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.