Question

Calcium carbonate reacts with aqueous $HCl$ to give $CaC{l_2}$ and $C{O_2}$ according to the reaction-
$CaC{O_3}(s) + 2HCl(aq) \to CaC{l_2}(aq) + C{O_2}(g) + {H_2}O(l)$
The mass of $CaCO{}_3$ 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

Hint: We would use the concept of stoichiometry as an approach to solve the given question. The concept of stoichiometry is defined as the concept in which the relation between quantities of reactant and product is estimated when the amounts of separate reactants are known.

Complete step by step answer:
As we know molarity means the number of moles of solute is calculated in 1000 ml of solution and given 0.75 M $HCl$ means 0.75 moles of $HCl$ are present in 1000 ml of solution.
So, 1000 ml of solution contain =0.75 moles of $HCl$
And, 1 ml of solution contain= $\dfrac{{0.75}}{{1000}}$moles of $HCl$
Then 25 ml of solution contain=$\dfrac{{0.75 \times 25}}{{1000}}$= 0.01875 mol
As from the chemical equation given, we know
2 mol of $HCl$ reacts with = 1 mol of $CaCO{}_3$
1 mol of $HCl$ reacts with =$\dfrac{1}{2}$mol of $CaCO{}_3$
0.01875 mol of $HCl$ reacts with = $\dfrac{1}{2} \times 0.01875 = 0.009375$mol
Now calculate the molar mass of $CaCO{}_3$. Since the mass of calcium is 40 g and the mass of carbon and oxygen is 12 and 16 gram respectively. On adding the mass of each atom present in $CaCO{}_3$ we get:
$
  40 + 12 + 3 \times 16 \\
   \Rightarrow 40 + 12 + 48 = 100g \\
$
Mass of 1 mol of $CaCO{}_3$=100g
Then the mass of 0.009375 mol of $CaCO{}_3$=$100 \times 0.009375 = 0.9375g$
So, the correct answer is “Option B”.

Note: Chemical reactions are symbolically represented by chemical equations with reacting material written on left and resultant product on the right. The number written in front of entities, such as ions, atoms or molecules taking part in a reaction is known as stoichiometric coefficient and they are used to balance a chemical reaction.