Please please please help me with this simple chemistry problem!

suppose a solution containing 3.50 g of sodium phosphate is mixed with a solution containing 6.40 g of barium nitrate. how many grams of barium phosphate can be formed?

2Na3PO4 + 3Ba(NO3)2 -------------------> 6NaNO3 + Ba3(PO4)2
3.50g ----------6.40g-------------------------…

for 2Na3PO4

3.50g/227m.mass = .01 moles

for 3Ba(NO3)2

6.4g/987 m.mass =.006 moles

since 2Na3PO4 is in excess... you use the limiting which is 3Ba(NO3)2 at 0.006 moles

since 3Ba(NO3)2 has 3x the moles Ba3(PO4)2 has you divide the number of moles by 3
which is .002

the molar mass of Ba3(PO4)2 is 601
g = moles X molar mass

so

1.2g of barium phosphate can be formed...



that was a hell of a problem.
You're welcome