Asample of 30.00 grams of potassium phosphate was added to a solution containing 100.0 ml of 0.25 m calcium nitrate. calcium phosphate was precipitated according to the following reaction. 2k3po4(s) + 3ca(no3)2(aq) - ca3(po4)2(s) + 6kno3(aq) 2.25 grams of calcium phosphate were recovered. calculate the theoretical yield and the percent yield.

Theoretical yield of Ca₃(PO₄)₂ = 19.67 g

Percent yield of Ca₃(PO₄)₂ = 11.4 %

Explanation:

Data given

Potassium Phosphate  K₃PO₄  = 30.00 g

calcium nitrate Ca(NO₃)₂ of 0.25 M  = 100.0 mL

Reaction given:

        2K₃PO₄(s) + 3Ca(NO₃)₂(aq) >  Ca₃(PO₄)₂(s) + 6KNO₃(aq)

Solution :

Part A :

To find the theoretical yield of Calcium phosphate

Look at the reaction :

             2K₃PO₄(s) + 3Ca(NO₃)₂(aq) > Ca₃(PO₄)₂(s) + 6KNO₃(aq)

               2 mol          3 mol                               1 mol                6mol

 So,

2 mol of  2K₃PO₄ give 1 mole of Ca₃(PO₄)₂

if we represent mole in mass then

Molar Mass of K₃PO₄ =  3(39) + 31 + 4(16)

Molar Mass of K₃PO₄ =  212 g/mol

and

Molar mass of Ca₃(PO₄)₂  =  [ 3(40) + 2((31) + 4(16))]

Molar mass of Ca₃(PO₄)₂  =  [ 120 + 62 + 96] = 278 g/mol

Molar mass of Ca₃(PO₄)₂ = 278 g/mol

Now, if we write it in grams

  2K₃PO₄(s)    +      3Ca(NO₃)₂(aq) > Ca₃(PO₄)₂(s)    +     6KNO₃(aq)

   2 mol                    3 mol                               1 mol                      6mol

  2 mol ( 212 g/mol)                                         1 mol (278 g/mol)

   424 g                                                               278 g

So from the above information we come to know that

    424 g of K₃PO₄ gives 278 g of Ca₃(PO₄)₂

then 30 g will give how much Ca₃(PO₄)₂ Calcium phosphate

Apply unity formula

    424 g of K₃PO₄ ≅ 278 g of Ca₃(PO₄)₂

       30 g of K₃PO₄ ≅ x g of Ca₃(PO₄)₂

By doing cross multiplication and rearranging the above values

      x g of Ca₃(PO₄)₂ = 278 g x  30 g / 424 g

      x g of Ca₃(PO₄)₂ = 19.67 g

So,

30 g of K₃PO₄ will give 19.67 g Ca₃(PO₄)₂ theoretically.

Theoretical yield of Ca₃(PO₄)₂ = 19.67 g

Now,

Part B :

To find percent yield of Ca₃(PO₄)₂

Solution

Data given:

       Actual yield of Ca₃(PO₄)₂ = 2.25 g

       Theoretical yield of Ca₃(PO₄)₂ = 19.67 g

Formula to be used:

         Percentage yield = Actual yield/ theoretical yield x 100 %

Put values in Above formula

       Percent yield = 2.25 g / 19.67 g x 100 %

       Percent yield =  0.1144 / 100 %

      Percent yield = 11.4 %

So,

Percent yield of Ca₃(PO₄)₂ = 11.4 %

explanation:

detials on each model with digrams

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons