Answer to Question #108017 in Chemistry for elle

Question

Consider the reaction 3CaCl2(aq) + 2Na3PO4(aq) --> Ca3(PO4)2(s) + 6 NaCl(aq) (molar mass of calcium phosphate is 310.18 g/mol)

What mass of Ca3(PO4)2 is produced from 94.2 mL of 0.250 M Na3PO4 solution and 275 mL of 0.150 M

Answer 1

3CaCl_2(aq) + 2Na₃PO_4(aq) --> Ca₃(PO₄)_2(s) + 6NaCl_(aq)

The number of moles of CaCl₂ used in the reaction equals:

n(CaCl_2(aq)) = C(CaCl_2(aq)) × М(CaCl_2(aq)) = 0.250 M × 0.0942 L = 0.02355 mol

The number of moles of Na₃PO₄ used in the reaction equals:

n(Na₃PO_4(aq)) = C(Na₃PO_4(aq)) × М(Na₃PO_4(aq)) = 0.150 M × 0.275 mL = 0.04125 mol

As n(CaCl_2(aq)) < n(Na₃PO_4(aq)), calcium chloride is a limiting reactant.

From here, mass of Ca₃(PO₄)₂ produced in the reaction equals:

m(Ca₃(PO₄)₂) = [n(CaCl₂)/2] × Mr(Ca₃(PO₄)₂) = [0.0235 mol / 2] × 310.18 g/mol = 3.64 g

Answer: 3.64 g