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?

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.

Sorry i dont understand the question can you rephase it?