Posted by Kerry on Monday, March 8, 2010 at 9:49am.
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.
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.
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.
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?
Na3PO4 + Ba(NO3)2 = Ba(PO4)2 + NaNO3
64
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