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Question: If \[0.5\] moles of \[BaC{l_2}\] ​ is mixed with \[0.2\] moles of \[N{a_3}P{O_4}\] ​, the maximum nu...

If 0.50.5 moles of BaCl2BaC{l_2} ​ is mixed with 0.20.2 moles of Na3PO4N{a_3}P{O_4} ​, the maximum number of moles of Ba3(PO4)2B{a_3}{(P{O_4})_2}​ that can be formed is:
(A) 0.10.1
(B) 0.20.2
(C) 0.50.5
(D) 0.70.7

Explanation

Solution

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 Ba3(PO4)2B{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 BaCl2BaC{l_2} with Na3PO4N{a_3}P{O_4} is:
3BaCl2+2Na3PO4Ba3(PO4)2+6NaCl3BaC{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 BaCl2BaC{l_2} required to form 11 mole of Ba3(PO4)2=3B{a_3}{(P{O_4})_2} = 3
Number of moles of Na3PO4N{a_3}P{O_4} required to form 11 mole of Ba3(PO4)2=2B{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 = 0.2
Now we will analyse this data to find the limiting reagent in the reaction.

| Moles of BaCl2BaC{l_2}| Moles of Na3PO4N{a_3}P{O_4}
---|---|---
Initial| 33| 22
Final| 0.50.5| 0.20.2

As you can see from the table, 33 moles of BaCl2BaC{l_2} reacts with 22 moles of Na3PO4N{a_3}P{O_4} to give 11 mole of Ba3(PO4)2B{a_3}{(P{O_4})_2}. Now, we will calculate the number of moles of Na3PO4N{a_3}P{O_4} required to react with 0.50.5 moles of BaCl2BaC{l_2}. They will be:
0.50.5 moles of BaCl2BaC{l_2} will react with =0.53×2=0.33moles = \dfrac{{0.5}}{3} \times 2 = 0.33\,moles
The number of moles of Na3PO4N{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 0.20.2 moles of Na3PO4N{a_3}P{O_4}.
Number of moles of Ba3(PO4)2B{a_3}{(P{O_4})_2} formed when 22 mole of Na3PO4N{a_3}P{O_4} reacts =1 = 1
Number of moles of Ba3(PO4)2B{a_3}{(P{O_4})_2} formed when 11 mole of Na3PO4N{a_3}P{O_4} reacts =12 = \dfrac{1}{2}
Number of moles of Ba3(PO4)2B{a_3}{(P{O_4})_2} formed when 0.20.2 mole of Na3PO4N{a_3}P{O_4} reacts =0.22=0.1 = \dfrac{{0.2}}{2} = 0.1
Hence, the total number of moles of product formed will be 0.10.1.
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.