Question

Calcium carbonate reacts with aqueous HCl to give ${\text{CaC}}{{\text{l}}_{\text{2}}}$ and ${\text{C}}{{\text{O}}_{\text{2}}}$ according to the reaction -${\text{CaC}}{{\text{O}}_{\text{3}}}{\text{(s) + 2HCl(aq)}} \to {\text{CaC}}{{\text{l}}_{\text{2}}}{\text{(aq) + C}}{{\text{O}}_{\text{2}}}{\text{(g) + }}{{\text{H}}_{\text{2}}}{\text{O(l)}}$. The mass of ${\text{CaC}}{{\text{O}}_{\text{3}}}$ which is required to react completely with 25 ml of 0.75 M HCl is
A.1.825g
B.0.9375g
C.1.8357g
D.0.46875g

Answer 1

1 mole of ${\text{CaC}}{{\text{O}}_{\text{3}}}$ is reacting with 2 moles of HCl. 1 mole of ${\text{CaC}}{{\text{O}}_{\text{3}}}$ contains 100g ${\text{Ca(40) + C(12) + }}{{\text{O}}_{\text{3}}}{\text{(16 X 3)}}$ and 1 mol of HCl contains 36.5 g, the mass of Cl is 35.5 and that of H is 1g. Molarity is the number of moles of solute dissolved in a unit litres of the solution.

Complete step by step answer:
Here, the volume of HCl given is 25 ml.
The molarity of 25 ml of HCl is 0.75 M
We know that the molar mass of HCl is 36.5g which implies that 36.5g HCl is present in 1 mol of HCl. So, we can find out the grams of HCl present in 0.75 M.
$1{\text{ mol HCl}} \to {\text{36}}{\text{.5g}}$
$0.75{\text{ mol HCl}} \to ({\text{36}}{\text{.5}} X {\text{0}}{\text{.75)g}}$
As we have already discussed the equation for molarity.
${\text{Molarity = }}\dfrac{{{\text{No of moles of solute}}}}{{{\text{Volume of solution in L}}}}$
Therefore, in 1L, $36.5 X 0.75$g of HCl is present
So, the mass of HCl in 1000 mL is $\dfrac{{36.5 X 0.75}}{{1000}}$ grams
In the given question, the volume of HCl is given as 25mL.
So, the mass of HCl in 25mL is $\dfrac{{36.5 X 0.75}}{{1000}} X 25$
${\text{Mass of HCl in 25ml = }}\dfrac{{{\text{36}}{\text{.5 X 0}}{\text{.75}}}}{{{\text{1000}}}}{\text{ X 25}}$ = 0.684g
Mass of 0.75M HCl in 25 mL is 0.684g
From the balanced reaction given in the question, we can observe that 1 mol ${\text{CaC}}{{\text{O}}_{\text{3}}}$ is reacting with 2 moles of HCl.
$2{\text{ mol HCl}} \to {\text{1 mol CaC}}{{\text{O}}_3}$
$(2 X 36.5)g{\text{ HCl}} \to {\text{100g CaC}}{{\text{O}}_3}$
$73g{\text{ HCl}} \to {\text{100g CaC}}{{\text{O}}_3}$
So we can say that 73 g of HCl is reacting with 100g of ${\text{CaC}}{{\text{O}}_{\text{3}}}$.
We already found out from the given volume and molarity of HCl that the mass of HCl reacting is 0.684g.
$73g{\text{ HCl}} \to {\text{100g CaC}}{{\text{O}}_3} \\\ 0.684g{\text{ HCl}} \to '{\text{x' g CaC}}{{\text{O}}_3} \\\$
The above step means that 0.684g of HCl is reacting an unknown mass of ${\text{CaC}}{{\text{O}}_{\text{3}}}$
So, it is given ‘x’
${\text{x = }}\dfrac{{{\text{100 X 0}}{\text{.684}}}}{{{\text{73}}}}$
${\text{x = 0}}{\text{.9375g}}$
The mass of ${\text{CaC}}{{\text{O}}_{\text{3}}}$ which is required to react completely with 25 ml of 0.75 M HCl is 0.9375g
Thus, the correct option is (B).

Note: There is no specific equation to find the answer to such questions. However, a systematic way of doing may help to get the answer. Sometimes, the question can be to find the mass of any product that is forming. The same method can be used. Make sure that the reaction given is balanced. If it is not balanced; balance the reaction by finding out the number of moles reacting or the number of moles of the product obtained.