What is the maximum mass of the precipitate formed when 127 mL of 2.1 M potassium phosphate reacts with magnesium chloride?

Question

2 K_3PO_4 (aq) + 3 MgCl_2 (aq) = Mg_3(PO_4)_2 (s) + 6 KCl (aq)

Answer 1

2 K_3PO_4 (aq) + 3 MgCl_2 (aq) = Mg_3(PO_4)_2 (s) + 6 KCl (aq)

127 mL K3PO4 x (1L/1000mL) x (2.1 mol K3PO4/1 L K3PO4) x (1 mol Mg3(PO4)2/2 mol K3PO4) x (262.86 g Mg3(PO4)2/1 mol Mg3(PO4)2 = 35.05 g Mg3(PO4)2

check my math...