Question

If $0.7$ mole of barium chloride is treated with $0.4$ mole of potassium sulphate, the number of moles of barium sulphate formed are:
A.$0.7$
B.$0.4$
C.$0.35$
D.$0.2$

Answer 1

Hint: To solve this question, you must recall basic stoichiometric fundamentals. Stoichiometry is based on the law of conservation of mass which suggests that the total mass of reactants is equal to the total mass of products. Thus, if the amount of separate reactants are known to us, then the amount of products can be determined.

Complete step by step answer:
For the given question, the reaction taking place is,
${\text{BaC}}{{\text{l}}_{\text{2}}} + {{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to {\text{BaS}}{{\text{O}}_{\text{4}}} + {\text{2KCl}}$
One mole of barium chloride reacts with one mole of potassium sulphate and gives one mole of barium sulphate.
We know that both the reactants react in the same proportions. Since, the amount of potassium sulphate present is less than the amount of barium chloride present, we can say that potassium sulphate is the limiting agent.
In other words, only $0.4$ moles of the reactants will react and $0.3$ moles of barium chloride will be left as extra.
From this, we can also infer that the amount of barium sulphate produced will also be $0.4$ moles.

Thus the correct answer is B.

Note:
In general, chemicals combine in definite ratios in the chemical reactions. Since chemical reactions can neither create nor destroy matter, nor transmute one element into another, thus the amount of each element must be the same throughout the entire reaction. For instance, the number of atoms of a given element X on the reactant side must be equal to the number of atoms of that element on the product side, irrespective of whether or not all of those atoms are involved in a reaction.