Question

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)

Answer 1

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.

Answer 2

Thanks for your help DrBob but the equation given is the one my teacher gave us. I just balanced it and I am not sure where I went wrong. I understand how to do the rest of the problem. I just need help to balance the equation.

Answer 3

<b>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.</b>

Suppose a solution containing 3.50g of Na3PO4 is mixed with an excess of Ba(NO3)2. how many grams of Ba2(PO4)2 <b> This should be Ba3(PO4)2</b> can be formed.

2-Na3PO4 +3-Ba3(NO3)2= 1-Ba3(PO4)2 + 6-NaNO3.(Balanced equation)
<b>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.

</b>

Answer 4

If 3.50g of Na3Po4 and 6.40g of Ba(NO3)2 are added together how many grams of Ba3(PO4)2 will be produce? which reactant is the limiting reactant?

Answer 5

Na3PO4 + Ba(NO3)2 = Ba(PO4)2 + NaNO3

Answer 6

64

Answer 7

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