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?

Question

Felicity Benson

Chemistry

2 September, 20:23

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?

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Genevieve Clarke · Answer 1

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.