Lithium phosphate (Li3PO4) and barium chloride (BaCl2) react in an ionic precipi

Question

Lithium phosphate (Li3PO4) and barium chloride (BaCl2) react in an ionic precipitation reaction to form barium phosphate and lithium chloride, according to the following: 2LiPO4(aq) + 3BaCl2(aq) --> Ba3PO4(s) + 6LiCl(aq)   DeltaH=14kj.
a) what volume (in mL) of 0.16 M Li3PO4 is required to react exactly with .548 mL of 0.33 M BaCl2? show work
b) What is the theoretical yield (in moles) of Ba3(PO4)2? show work
c) When we assume that the reaction goes to completion, what assumption are we making about the numerical value of the Keq for this reaction? Explain reasoning and give an example of a possible value for Keq for a complete reaction.
d) if the reaction did not go to completion, and one wated to shift the equiibrium more to the right in order to produce more product, name one change to reaction conditions that would accomplish this purpose.

Explanation / Answer

2LiPO4(aq) + 3BaCl2(aq) --> Ba3PO4(s) + 6LiCl(aq)   DeltaH=14kj.

a)2 moles of the lithium compound reacts with three moles of Barium comound.

hence,

3*0.16*v=0.548*0.33*2

or v=0.7535 mL

b)number of moles of barium phosphate = 0.5*0.16*0.7535

=0.06028 moles

c)when we say the the reaction goes to completion, we mean that the Keq for the reaciton is infinity since only then the concentration of the products will be zero. so possible Keq= infinity

d) if we wanted to shift the equation to the right, adding more products would have helped since it would have lowered the Q of the reaction. Removing the formed product also shifts the reaction to the right.

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