How many grams of solid ca3(po4)2 can be formed from a reaction of 51.4 ml of 0.758 m cacl2 and 21.5 ml of 1.04 m na3po4?

Answer is: 3,41 g of Ca₃(PO₄)₂.
Chemical reaction: 3CaCl₂ + 2Na₃PO₄ → Ca₃(PO₄)₂ + 6NaCl.
V(CaCl₂) = 51,4 mL = 0,0514 L.
c(CaCl₂) = 0,758 mol/L.
n(CaCl₂) = V(CaCl₂) · c(CaCl₂).
n(CaCl₂) = 0,0514 L · 0,758 mol/L.
n(CaCl₂) = 0,039 mol.
V(Na₃PO₄) = 21,5 mL = 0,0215 L.
c(Na₃PO₄) = 1,04 mol/L.
n(Na₃PO₄) = 0,0215 L · 1,04 mol/L.
n(Na₃PO₄) = 0,022 mol, limiting reagens.
From chemical reaction: n(Na₃PO₄) : n(Ca₃(PO₄)₂) = 2 : 1.
n(Ca₃(PO₄)₂) = 0,011 mol.
m(Ca₃(PO₄)₂) = 0,011 mol · 310,17 g/mol.
m(Ca₃(PO₄)₂) = 3,41 g.