contestada

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?

Respuesta :

Ishankahps
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.

Otras preguntas