Finding Mass of Ba3(PO4)2 with Na3PO4 & Ba(NO3)2

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of the mass of barium phosphate (Ba3(PO4)2) formed when sodium phosphate (Na3PO4) is mixed with barium nitrate (Ba(NO3)2). The participants explore the stoichiometry of the reaction and the correct application of molar masses in their calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a calculation for the mass of Ba3(PO4)2 based on the given mass of Na3PO4 and the stoichiometric equation.
  • Another participant challenges the initial calculation, pointing out the need to consider the stoichiometry of the reaction, specifically the coefficient of sodium phosphate.
  • A later reply acknowledges the oversight regarding the coefficient and confirms the initial calculation after correcting for it.
  • There is a suggestion that the molar mass of Na3PO4 should be adjusted in the calculations due to its stoichiometric coefficient.
  • Participants express uncertainty about the correctness of the molar mass used and the method applied.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the final mass calculation, as there are differing views on the correct application of stoichiometry and molar masses. Some participants agree on the correction regarding the coefficient, while others express uncertainty about the calculations.

Contextual Notes

There are unresolved issues regarding the correct molar mass to use in the calculations, as well as the implications of the stoichiometric coefficients in the balanced equation.

Who May Find This Useful

Students working on stoichiometry problems in chemistry, particularly those involving precipitation reactions and mass calculations.

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 ?
 
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Please tell me if I'm right!
 
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.
 
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
 
Not quite. You notice that there are two phosphates per mole in barium phosphate?
 
yes and i think I'm not wrong in calculating the molar mass which is 601.93, but is that method correct?
 
OOps! My bad. I didn't notice the 2 in front of the sodium phosphate!

You got it.
 
YAY :D i was really starting that today is not my day..
so m=1505.51?
 
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
Please reply to my posts!
 
  • #11
Chawki. You're correct.
 
  • #12
jtabije said:
Chawki. You're correct.

Thank you jtabije.
 

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