Suppose a solution containing 3.50g of Na3PO4 is mixed with an excess of Ba(NO3)2. how many grams of Ba2(PO4)2 can be formed.
2-Na3PO4 +3-Ba3(NO3)2= 1-Ba3(PO4)2 + 6-NaNO3.(Balanced equation)
Your equation is not correct AND it isn't balanced either.
1. Start by writing a balanced equation.
2. Convert 3.50 g Na3PO4 to moles. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles of Na3PO4 to moles of Ba3(PO4)2.
4. Now convert moles Ba3(PO4)2 to grams. g = moles x molar mass.
Post your work if you get stuck.
Perhaps you just copied the equation incorrectly to your notes or you copied it to the post incorrectly. Here is the problem as you wrote it; my comments are in bold at appropriate places.
Suppose a solution containing 3.50g of Na3PO4 is mixed with an excess of Ba(NO3)2. how many grams of Ba2(PO4)2 This should be Ba3(PO4)2 can be formed.
2-Na3PO4 +3-Ba3(NO3)2= 1-Ba3(PO4)2 + 6-NaNO3.(Balanced equation)
Na3PO4 is OK. Barium nitrate is Ba(NO3)2. Ba3(PO4)2 is OK; I suspect the incorrect formula in the problem part is a typo. NaNO3 is OK.
Correct equation:
Na3PO4 + Ba(NO3)2 ==>Ba3(PO4)2 + NaNO3
Correct balancing:
2Na3PO4 + 3Ba(NO3)2 ==> Ba3(PO4)2 + 6NaNO3. In your "balanced" equation, the Ba doesn't balance, mostly because the formula for barium nitrate, Ba(NO3)3, is wrong. You should be able to see that you have 9 Ba atoms on the left and only 3 on the right. Perhaps the 3 is just a typo BUT you should catch those when you check them. ALWAYS check them.