Question

If 40 g of \[CaC{O_3}\] is treated with 40 g of HCl, which of the reactants will act as limiting reagent?
(A) \[CaC{O_3}\]
(B) HCl
(C) Both are equal
(D) Cannot be calculated

Answer 1

Hint: Try to recall that the amount of product formed depends upon the reactant which has reacted completely. Now, by using this you can easily find the correct option from the given ones.
Complete step by step solution:
It is known to you that the amount of product formed then depends upon the reactant which has reacted completely. This reactant which reacts completely in the reaction is called the limiting reactant or limiting reagent.
The reactant which is not consumed completely in the reaction is called excess reactant as the excess of this reactant is left unreacted.
Calculation:
The balanced chemical equation for the above given reactants: \[CaC{O_3} + 2HCl \to CaC{l_2} + {H_2}O + C{O_2}\]
Step 1. To convert the given amounts into moles
Mass of \[CaC{O_3}\] given= 40g
Molar mass of \[CaC{O_3}\]=100
\[So,number{\text{ of moles of CaC}}{{\text{O}}_3} = \frac{{mass{\text{ of CaC}}{{\text{O}}_3}}}{{Molar{\text{ }}mass{\text{ }}of{\text{CaC}}{{\text{O}}_3}}} = \frac{{40}}{{100}} = 0.4\]
Mass of HCl given= 40g
Molar mass of HCl=36.5
\[So,number{\text{ of moles of HCl}} = \frac{{mass{\text{ of HCl}}}}{{Molar{\text{ }}mass{\text{ }}of{\text{ HCl}}}} = \frac{{40}}{{36.5}} = 1.1\].
Step 2: To identify the limiting reagent
From the above balanced equation, 1 mole of \[CaC{O_3}\] reacts with 2 mole of HCl
So, 0.4 mole of \[CaC{O_3}\] will react with HCl= \[ = 2 \times 0.4 = 0.8\]mol
But we have 1.1 mole of HCl. Hence, \[CaC{O_3}\] is the limiting reagent and HCl is the excess reagent.

Therefore, from above we can easily conclude that option A is the correct option to the given question.

Note:
- It should be remembered that limiting reactant or reagent decides the amounts of other reactants reacted or the amount of products formed.
- Also, you should remember that the actual yield of a product is usually less than the theoretical yield because of certain side reactions taking place.