Finding the mass

  • Thread starter chawki
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  • #1
chawki
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Homework Statement


When solutions of sodium phosfate and barium nitrate are mixed, a precipitate of barium
phospfate is formed according to the equation
2 Na3PO4(aq) + 3 Ba(NO3)2(aq) ---> Ba3(PO4)2(s) + 6 NaNO3(aq)

Homework Equations


What mass of Ba3(PO4)2 is formed when a solution containing 410 g of Na3PO4 is mixed with a solution containing an excess of Ba(NO3)2?

The Attempt at a Solution


is it 1505.1g ?
 

Answers and Replies

  • #2
chawki
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Please tell me if i'm right!
 
  • #3
chemisttree
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NO! Don't force us to look up the data. ITS YOUR WORK, NOT OURS!

You've probably put in the proper effort but you haven't demonstrated that to us.
 
  • #4
chawki
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ok sorry for that...i'm new on this
well.. i used an old method from college..

2 Na3PO4(aq) + 3 Ba(NO3)2(aq) ---> Ba3(PO4)2(s) + 6 NaNO3(aq)

410g-------------------------------------m
163.97g/mol-----------------------------601.93g/mol

m=(410*601.93)/163.97
m=1505.1g
 
  • #5
chemisttree
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Not quite. You notice that there are two phosphates per mole in barium phosphate?
 
  • #6
chawki
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yes and i think i'm not wrong in calculating the molar mass which is 601.93, but is that method correct?
 
  • #7
chemisttree
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OOps! My bad. I didn't notice the 2 in front of the sodium phosphate!

You got it.
 
  • #8
chawki
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YAY :D i was really starting that today is not my day..
so m=1505.51?
 
  • #9
chawki
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Maybe i was wrong about the molar mass of Na3PO4 that we should write when using that method, because we have 2 Na3PO4, so i think we should write 2*163.97

2 Na3PO4(aq) + 3 Ba(NO3)2(aq) ---> Ba3(PO4)2(s) + 6 NaNO3(aq)

410g-------------------------------------m
2*163.97g/mol---------------------------601.93g/mol

and thus we will get m=752.55g
 
  • #10
chawki
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Please reply to my posts!
 
  • #11
jtabije
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Chawki. You're correct.
 
  • #12
chawki
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