Question

100 g sample of calcicum carbonate is reaction with 70g of orthophosphoric acid. Calculate
(a) the number of grams of calcium phosphate that could be produced.
(b) the number of grams of excess reagent that will remain unreacted.

Answer 1

The balanced equation is
`underset("=300g")underset(3(40+12+48))underset("3mol")(3CaCO_(3))+underset(=196g)underset(2(3+3+64))underset("2mol")(2H_(3)PO_(3)) to underset(=310g)underset((3xx40+2xx64))underset("1 mol")(Ca_(3)(PO_(4))_(2))+3CO_(2)+3H_(2)O`
`300"g of"CaCO_(3) "produce" Ca_(3)(PO_(4))_(2)=310"g or 1 mol"`
`100"g of"CaCO_(3) "would produce"`
`Ca_(3)(PO_(4))_(2)=(310)/(300)xx100`
=103g
=0.03mol
`196"g of "H_(3)PO_(4) "produce" Ca(PO_(4))_(2)=310"g of 1 mol"`
`70"g of "H_(3)PO_(4) "produce" Ca(PO_(4))_(2)=(310)/(196)xx70`
`=110.7g or 0.356"mol"`
The above values suggest that `CaCO_(3)` is the limiting reagent. Hence, calcium phosphate formed is 103g of 0.33 mole.
(b) For producing 103 g of `Ca_(3)(PO_(4))_(2), H_(3)PO_(4)` required will be `=(196)/(310)xx1033=65.12g`
Mass of remaining `H_(3)PO_(4)=(70-65.12)=4.88g`