This is a mulit step problem
How many grams of barium phosphate are produced from the reaction of 3.50g of sodium phosphate and 6.4g of barium nitrate? The other product is sodium nitrate.
1. Write the balanced equation for the reaction
2. How many grams of barium phosphate are produced
3. What is the excess reagent and how much is left over after the reaction?
4. When you do the reaction in the lab, you are able to isolate 4.00g of barium phosphate. What is the percent yield?
We need to know how much of this you know how to do yourself.Write the equation. Convert grams of what you have (both of what you have) to mols. Remember mols = g/molar mass.
Then use the equation to convert mols of what you have to mols of product. You will get two answers. Pick the smaller answer and that material will be the limiting reagent. Use that to calculate how much of the other reagent remains. Then percent yield = (4.00/theoretical)x100. Theoretical is the amount you calculated above; i.e., the smaller value you chose that determined the limiting reagent. Post you work if you get stuck.
I am having trouble from the get go because I don't know if my balanced equation is right. My teacher has only taught us how to do this with one number (3.50g), not two (3.50g and 6.4g).
NA(PO4)3 up top + BA 2 up top (NO3) - BA 2 up top (PO4) 3 up top + NA(NO3)
Is that right, it does not look right and if so, everything already balances out, which confuses me for the next step. Would their only be one gram of barium phosphate? I know how to do the steps, I just don't know how to do this with two number.
2Na3PO4 + 3Ba(NO3)2==>6NaNO3 + Ba3(PO4)2
Thanks for helping, I appreciate it. I assume that Ba3 is a little 3?
So since the next question is how many grams of barium phosphate are produced? I see this as 1Ba and then do the rest of my work?
I am sorry for the delay in answering but I had a theater date and had to go before I could finish with your question. Yes, the 3 following Ba is a small 3 and it is a subscript. The 3 in front of Ba is a coefficient and is a large 3.
Here is the equation again.
3Na3PO4 + 3Ba(NO3)2 =>6NaNO3 + Ba3(PO4)2
Step 2. Convert what you have into moles. mols = grams/molar mass.
a. mols Na3PO4 = 3.50/molar mass = ?? mols Na3PO4.
b. mols Ba(NO3)2 = 6.4/molar mass = xx mols Ba(NO3)2.
Step 3. Using the coefficients in the balanced equation, convert mols of what you have (2a and 2b) into what you want (in this case Ba3(PO4)2.
a. from Na3PO4. ??mols Na3PO4 x (1 mol Ba3(PO4)2/2 mols Na3PO4) = yy mols Ba3(PO4)2.
b. from Ba(NO3)2. xxmols Ba(NO3)2 x (1 mol Ba3(PO4)2/3 mols Ba(NO3)2) = zz mols Ba3PO4)2.
c. You have two answers for mols Ba3(PO4)2 produced and obviously both can't be right. The correct one is the smaller of the two (either from 2a or from 2b which converts into 3a and 3b). That one (either 2a or 2b) will be the limiting reagent and ALL of it will be consumed. The other one will have some consumed but some of it will remain. We will come back to it later. Note that the units in the conversion cancel from mols Na3PO4 or Ba(NO3)2 and convert into mols Ba3(PO4)2 by using the coefficients in the balanced equation.
Step 4. Using the answer from 3c, convert mols of the appropriate (either yy or zz) into grams Ba3(PO4)2 by multiplying by molar mass of Ba3(PO4)2.
Step 5. Then percent yield is
% yield = [4.0g/g Ba3(PO4)2 from step 4]x100 = cc
Now--that leaves just one more part of the problem which is part 3 from your original post. This becomes a simple stoichiometry problem. You have the limiting reagent which is the one that uses ALL of the starting material. Use that (you have the mols) to convert (using the equation as above) to mols of the excess material. Whatever number you obtain will be less that the starting mols. Convert to grams, subtract from the starting grams and that will be the amount remaining.
I hope this helps.
just wondering if Ba(NO3)2 is insoluble or soluble in water at 25degreesC. Thank you