Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.
Explanation : Given,
Mass of
= 12.00 g
Mass of
= 10.0 g
Molar mass of
= 164 g/mol
Molar mass of
= 111 g/mol
Molar mass of
= 58.5 g/mol
Molar mass of
= 310 g/mol
First we have to calculate the moles of
and
.


and,


Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is:

From the balanced reaction we conclude that
As, 3 mole of
react with 2 mole of 
So, 0.0901 moles of
react with
moles of 
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
and 
From the reaction, we conclude that
As, 3 mole of
react to give 6 mole of 
So, 0.0901 mole of
react to give
mole of 
and,
As, 3 mole of
react to give 1 mole of 
So, 0.0901 mole of
react to give
mole of 
Now we have to calculate the mass of
and 


and,


Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.