# Finding the mass

## 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!

chemisttree
Homework Helper
Gold Member
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

chemisttree
Homework Helper
Gold Member
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?

chemisttree
Homework Helper
Gold Member
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