Chemistry Question Solution Stochiometry And Determining The Concentration?

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To determine the concentration of ions after mixing 600.0 mL of 2.00 mol/L barium nitrate with 100.0 mL of 2.50 mol/L sodium phosphate, the balanced reaction equation is 3Ba(NO3)2 + 2Na3PO4 = Ba3(PO4)2 (solid) + 6NaNO3. Calculating moles, 1.2 moles of barium nitrate and 0.25 moles of sodium phosphate are found, indicating sodium phosphate is the limiting reagent. The excess barium nitrate is calculated as 0.825 moles after accounting for the reaction. Finally, the total volume of the solution is 0.7 liters, allowing for the calculation of ion concentrations by dividing the moles of each ion by this total volume. This approach yields the final concentrations of all ions present in the solution.
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Homework Statement



What is the concentration of all ions after 600.0mL of 2.00 mol/L Barium nitrate is added to 100.0 mL of 2.50 mol/L sodium phosphate.


The Attempt at a Solution


I first wrote down the equation : 3Ba(No3)2 + 2Na3PO4 = Ba3(PO4)2 (solid) + 6NaNO3
I then found the moles using n = c x v.
moles of Ba(No3)2 n = 2 x 0.6L = 1.2 mols
moles of Na3PO4 n = 2.5 x 0.1 = 0.25 mols.

I than found the LR: 0.25 mols Na3PO4 x 3 mols Ba(NO3)2/2Na3PO4 = 0.375 mols Ba(NO3)2

So I have more than i need for barium nitrate so my limiting reagent is sodium phosphate.

Now what i am having trouble with is, is that should i find the excess reagent by doing 1.2 - 0.375 = 0.825 or do i have to find the mols of the solid barium phosphate and then find the limiting reagent using that.. Please help, thanks!
 
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anyone care to help??
 
You are provided with the concentrations and volumes of the reactants. Find out the moles of reactants and the final moles of all the ions as the question says. Then divide the moles by the total volume i.e. 0.7 litres.
 
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