Question

In an experiment, 22.00 mL of a 0.250 M calcium chloride are combined with 35.00 mL of a 0.200 M sodium phosphate solution to produce a white precipitate. After the precipitate is collected and dried, it is found to have a mass of 0.973 g. What is the percent yield of this reaction?

Answer 1

Percent yield = (Actual yield / Theoritical yield) x 100%

Actual yield is given as 0.973 g

Calculation of theoritical yield;

3CaCl₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6NaCl(aq)
Stoi. ratio 3 : 2 : 1
Initial moles 5.5 x 10⁻³ 7 x 10⁻³
Reacted 5.5 x 10⁻³ ( 5.5 x 10⁻³ x 2) / 3
Final moles - 3.33 x 10⁻³ (5.5 x 10⁻³) / 3

to calculate the moles following equation was used.
moles = volume (L) x Molarity ( mol L⁻¹)

The molar mass of Ca₃(PO₄)₂(s) = 310 g mol⁻¹

mass of Ca₃(PO₄)₂(s) = moles x molar mass
= (5.5 x 10⁻³) / 3 mol x 310 g mol⁻¹
= 0.568 g

percent yield = (0.973 g / 0.568 g) x 100%
= 171.30 %

Since the percent yield is over that 100, there is an error in actual yield.
the error can be in measuring instrument or the product is not dried well.