Chemistry problem of the day ...

First you must complete the reaction

2Na3PO4+3Ca(NO3)2->Ca3(PO4)2+6NaNO3

The determine the quantity of reactants (the left)
25ml*0.185moles/1000ml Na3PO4 (0.004625 mole)
34ml*0.140moles/1000ml Ca(NO3)2 ( 0.00476 mole)

Now determine the limiting reactant [Looks like Ca(NO3)2]

so (0.00476/3 (0.001587) moles of Ca3(PO4)2 and 6*0.001587 (0.00952) moles of NaNO3 will be formed, and 0.004625-0.00476*2/3 (0.001452) moles of Na3PO4 will be left.

And what is the total volume? its 59 (25+34)ml, right. And how many liters is this? Its 59/1000 (0.059 liter right?)

So the molar concentrations after reaction are
0.001587/0.059 M Ca3(PO4)2
0.00952/0.059 M NaNO3
and
0.001452/0.059 M Na3PO4
Right

But if I were you I'd check my arithmetic--

Rap

thank you for your reply. not the first time you've helped me out here. pardon me for not having stated my question clearly. I actually wanted to find the concentration of each ion after the reaction. thanks