Answer to Question #96123 in General Chemistry for Emily

Question #96123
BaCl2 + K3PO4 > Ba3(PO4)2 + KCl
Consider the unbalanced equation above. What volume of 0.320 M K3PO4 is needed to react with 2.85g of BaCl2?
1
Expert's answer
2019-10-08T08:19:39-0400

Solution.

First we equalize the chemical equation of the reaction:

"3BaCl2 + 2K3PO4=Ba3(PO4)2 + 6KCl"

Now find the amount of substance barium chloride:

"n(BaCl2)=\\frac{m}{M}"

n(BaCl2)=0.014 mole

"n(K3PO4)=\\frac{n(BaCl2) \\times 2}{3}"

n(K3PO4)=0.009 mole

"C(K3PO4)=\\frac{n}{V}"

"V(K3PO4)=\\frac{n}{C(K3PO4)}"

V(K3PO4)=28.13 ml

Answer:

V(K3PO4)=28.13 ml


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