Question

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?

Answer 1

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