Question

If you start with 1.91g of Na3PO4, what is the theoretical yield of Cu3(PO4)2 Molar mass: Na3PO4 is 380.12 g/mole, Cu3(PO4)2 is 434.60 g/mole, CuCl2 is 170.48 g/mole.

Answer 1

Assuming Na3PO4 as a limiting reagent. The balanced reaction is,

3CuCl2 + 2 Na3PO4 ----------> Cu3(PO4)2 + 6NaCl

Formula: moles = mass in g / molar mass in g/mol

Number of moles of Na3PO4 = 1.91g / 380.12 g/mol = 5.025×10^-3 moles

As per the equation, 2 moles of Na3PO4 gives 1 mole of Cu3(PO4)2

Thus,5.025×10^-3 moles of Na3PO4 gives = (5.025×10^-3) / 2 = 2.513 × 10^-3 moles of Cu3(PO4)2

2.513 × 10^-3 moles of Cu3(PO4)2 = 2.513 × 10^-3 mol × 434.60g/mol = 1.09 g

Thus theoretical yield = 1.09 g