Strontium chloride, SrCl2, reacts with sodium phosphate, Na3PO4 to form a precipitate according to the following reaction:

3 SrCl2 (aq) + 2 Na3PO4 (aq) -> Sr3(PO4)2 (s) + 6 NaCl(aq)

If 50.0 mL of 0.10 M SrCl2 is mixed with 50.0 mL or 0.10 M Na3PO4, how many grams of precipitate are formed?

ok, So I've set up the equation for the first part as:

?grams of Sr3(PO4)2 = 50 mL Na3PO4 x (1L/1000mL)(.10M Na3PO4/1 L Na3PO4)(163.94g Na3PO4/1 mol Na3PO4)(1 mol SrPO4/2 mol Na3PO4) = .40985 grams of SrPO4

Is this first part correct? I am studying for an exam tomorrow. Thank you so much!