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 %

b. is the answer hoped i        

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.