Mass of Ba3(PO4)2 precipitate from 410 g Na3PO4 and excess Ba(NO3)2

[chawki]

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 ?

 

[chawki]

Please tell me if I'm right!

 

[chemisttree]

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.

 

[chawki]

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

 

[chemisttree]

Not quite. You notice that there are two phosphates per mole in barium phosphate?

 

[chawki]

yes and i think I'm not wrong in calculating the molar mass which is 601.93, but is that method correct?

 

[chemisttree]

OOps! My bad. I didn't notice the 2 in front of the sodium phosphate!

You got it.

 

[chawki]

YAY :D i was really starting that today is not my day..
so m=1505.51?

 

[chawki]

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

 

[chawki]

Please reply to my posts!

 

[jtabije]

Chawki. You're correct.

 

[chawki]

Chawki. You're correct.

Thank you jtabije.