Question

If \[0.5\] moles of \[BaC{l_2}\] ​ is mixed with \[0.2\] 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: First we will write the complete balanced equation when Barium chloride reacts with sodium phosphate to give barium phosphate and sodium chloride salt. To calculate the maximum number of moles of \[B{a_3}{(P{O_4})_2}\]​ that can be formed we have to find the limiting reagent in the reaction.

Complete step by step answer:
The complete balanced equation of reaction of \[BaC{l_2}\] with \[N{a_3}P{O_4}\] is:
\[3BaC{l_2} + 2N{a_3}P{O_4} \to B{a_3}{(P{O_4})_2} + 6NaCl\]
From the equation we can deduce that,
Number of moles of \[BaC{l_2}\] required to form $1$ mole of $B{a_3}{(P{O_4})_2} = 3$
Number of moles of \[N{a_3}P{O_4}\] required to form $1$ mole of $B{a_3}{(P{O_4})_2} = 2$
Now, we are given, Total number of moles of \[BaC{l_2}\]$ = 0.5$
Total number of moles of \[N{a_3}P{O_4}\] $ = 0.2$
Now we will analyse this data to find the limiting reagent in the reaction.

	Moles of \[BaC{l_2}\]	Moles of \[N{a_3}P{O_4}\]
Initial	$3$	$2$
Final	$0.5$	$0.2$

As you can see from the table, $3$ moles of \[BaC{l_2}\] reacts with $2$ moles of \[N{a_3}P{O_4}\] to give $1$ mole of \[B{a_3}{(P{O_4})_2}\]. Now, we will calculate the number of moles of \[N{a_3}P{O_4}\] required to react with $0.5$ moles of \[BaC{l_2}\]. They will be:
$0.5$ moles of \[BaC{l_2}\] will react with $ = \dfrac{{0.5}}{3} \times 2 = 0.33\,moles$
The number of moles of \[N{a_3}P{O_4}\]is less than the given amount. Hence, it will be the limiting reagent. Now we will calculate the number of moles of product formed from $0.2$ moles of \[N{a_3}P{O_4}\].
Number of moles of \[B{a_3}{(P{O_4})_2}\] formed when $2$ mole of \[N{a_3}P{O_4}\] reacts $ = 1$
Number of moles of \[B{a_3}{(P{O_4})_2}\] formed when $1$ mole of \[N{a_3}P{O_4}\] reacts $ = \dfrac{1}{2}$
Number of moles of \[B{a_3}{(P{O_4})_2}\] formed when $0.2$ mole of \[N{a_3}P{O_4}\] reacts $ = \dfrac{{0.2}}{2} = 0.1$
Hence, the total number of moles of product formed will be $0.1$.
Therefore, option (A) is correct.

Note:
The limiting reagent in a chemical reaction is that reactant that is consumed completely after the reaction is completed. The amount of product that is formed is limited by this reagent, as the reaction cannot proceed further without the limiting reagent.