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.

Non metal is an element or substance that is not a metal metal is a solid material that is typically hard, shiny, malleable, fusible, and ductile, with good electrical and thermal conductivitymetalloid is an element (e.g., germanium or silicon) whose properties are intermediate between those of metals and solid nonmetals. they are electrical semiconductors.