Question

$ CaC{l_2}(aq) + N{a_2}C{O_3}(aq) \to CaC{O_3}(s) + 2NaCl(aq) $
 $ I $ . $ 2.50g $ of $ CaC{l_2} $ is fully disclosed in beaker of water and $ 2.50g $ of $ N{a_2}C{O_3} $ is fully dissolved in water in a second beaker. The two solutions are mixed to form a $ CaC{O_3} $ precipitate and aqueous $ NaCl $ . $ N{a_2}C{O_3} $ will be the limiting reactant in this experiment.
 $ II $ . Sodium carbonate has a molar mass which is less than the molar mass of calcium chloride.
(A) Statement $ I $ is true, Statement $ II $ is true.
(B) Statement $ I $ is true, statement $ II $ is false.
(C) Statement $ I $ is false, statement $ II $ is true.
(D) Statement $ I $ is false, statement $ II $ is false.

Answer 1

Hint: Limiting reagent in a reaction is defined by the compound that is in the lesser quantity as compared to the other reactant which gets left. In a specific reaction the reactants react to moles according to their stoichiometric coefficients. Moles are defined as the given mass divided by the molecular weight.

Complete answer:
First, we will write the reaction given above:
 $ CaC{l_2}(aq) + N{a_2}C{O_3}(aq) \to CaC{O_3}(s) + 2NaCl(aq) $
In this one mole of calcium chloride reacts with one mole of sodium carbonate to form one mole of calcium carbonate and two moles of sodium chloride.
Moles are defined as the given moles divided by the molecular weight.
Moles of calcium chloride
 $ moles = \dfrac{{given\,mass}}{{molecular\,weight}} $
 $ \Rightarrow moles = \dfrac{{2.50}}{{40 + 35.5 \times 2}} $
 $ \Rightarrow moles = \dfrac{{2.50}}{{111}} $
 $ \Rightarrow moles = 0.0225 $
Moles of sodium carbonate
 $ moles = \dfrac{{given\,mass}}{{molecular\,weight}} $
 $ \Rightarrow moles = \dfrac{{2.50}}{{23 \times 2 + 12 + 16 \times 3}} $
 $ \Rightarrow moles = \dfrac{{2.50}}{{46 + 12 + 48}} $
 $ \Rightarrow moles = \dfrac{{2.50}}{{106}} $
 $ \Rightarrow moles = 0.0235 $
Now as we know that one mole of calcium chloride reacts with one mole of sodium carbonate, therefore we will compare the moles of both the substances.
 $ CaC{l_2} $ is the limiting reagent in the reaction.
Statement $ I $ is false, statement $ II $ is false.Option (D) is correct.

Note:
The molecular weight is defined as the weight of one molecule of a compound the molecular weight is equal to the individual molecular weights in the multiplication of their coefficient in their lower cases.