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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