Question

If $0.5$ moles of $BaC{l_2}$ is mixed with $0.20$ moles of $N{a_3}P{O_4}$ , the maximum number of moles of $B{a_3}{(P{O_4})_2}$ that can be formed is:
(A) $0.1$
(B) $0.2$
(C) $0.5$
(D) $0.7$

Answer 1

Hint : We will write what is given in the question. We will write the equation and in the balanced form. Then the coefficients will tell the ratio of reactant and products. Then we will calculate the yield of $B{a_3}{(P{O_4})_2}$ after the reaction is completed. Let us write the step by step solution now.

Complete Step By Step Answer:
Step1. Moles of $BaC{l_2}$ given is $0.5$ moles.
Moles of given is $N{a_3}P{O_4}$ $0.20$ moles.
The new product that will be formed is $B{a_3}{(P{O_4})_2}$ .
Step2. The reaction will happen between Barium Chloride and Sodium Phosphate. The reaction will be a double substitution reaction. The reaction is given below.
$3BaC{l_2} + 2NaP{O_4} \to B{a_3}{(P{O_4})_2} + 6NaCl$
Step3. The coefficients of the balanced equation will give us the idea about the ratio of the reactants and the products that are formed. By seeing the reaction, we see that three moles of the $BaC{l_2}$ will react with two moles of $N{a_3}P{O_4}$ and produce one mole of the $B{a_3}{(P{O_4})_2}$ .
Step4. The ratio of the $BaC{l_2}$ : $N{a_3}P{O_4}$ will be $3:2$
$0.5$ mole of $BaC{l_2}$ needs $\frac{2}{3} \times 0.5 = 0.333$ of $N{a_3}P{O_4}$ . So it is the limiting agent and $BaC{l_2}$ is in excess amount.
So the $0.20$ moles of $N{a_3}P{O_4}$ will create $0.1$ of the $B{a_3}{(P{O_4})_2}$ . Because two mole of the $N{a_3}P{O_4}$ create the one mole of the $B{a_3}{(P{O_4})_2}$ .
Hence Option A is the correct option.

Note :
A double substitution reaction exchanges the cations or the anions of the two ionic compounds. It is also known as precipitation reaction sometimes when the product is solid precipitate. Often the solubility rules are used to predict if the double replacement reactions will occur or not.