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!