Question

What is the mass of calcium chloride formed when $2.5g$ of calcium carbonate is dissolved with excess hydrochloric acid? $(CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O)$

Answer 1

Hint:According to law of conservation of mass amount of calcium in reactant is equal to amount of calcium in product. Find the number of moles of calcium in calcium carbonate and number of moles of calcium in reactant and product are same means number of moles of calcium in calcium chloride is equal to number of calcium in calcium carbonate. And the number of moles is equal to the given mass divided by molar mass.

Complete step by step answer:
Given mass of calcium carbonate is $2.5g$.
Molar mass of calcium carbonate $(CaC{O_3})$ is given by
${M_{CaC{O_3}}} = 40 \times 1 + 12 \times 1 + 16 \times 3 = 100g$.
We know that the number of moles is equal to the given mass divided by molar mass.
Then number of moles of calcium in reactants is
${\text{Number of moles = }}\dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}} = \dfrac{{2.5}}{{100}} = 0.025moles$
Then the number of moles of calcium chloride $(CaC{l_2})$ is the number of moles of calcium that is ${\text{0}}{\text{.025 moles}}$.
Molar mass of calcium chloride is given by
${M_{CaC{l_2}}} = 40 + (35.5) \times 2 = 111g$
Then weight of calcium chloride is equal to number of moles and molar mass that is
${\text{Mass of CaC}}{{\text{l}}_2} = {\text{Number of moles}} \times {\text{Molar mass = 0}}{\text{.025}} \times {\text{111}}$
${\text{Mass of CaC}}{{\text{l}}_2} = 2.78g$

Then the mass of calcium chloride formed is $2.78g$ and the correct answer is option B.


Note: In a given reaction yield of formation of calcium chloride is 100% because excess hydrochloric acid is reacted with calcium carbonate. If we have given that yield is $x\% $ then the mass of the product is also $x\% $ of mass obtained when yield is 100%.