"Published in Newark, California, USA"
A 50.0 g sample of calcium carbonate is reacted with 35.0 g of phosphoric 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.
(Atomic Weights: Ca = 40, C = 12, O = 16, H = 1, and P = 31)
Solution:
The first thing that we need to do is to write the chemical equation of the reaction as follows
Balance the equation by putting the corresponding coefficients as follows
Molecular Weight of CaCO3:
Molecular Weight of H3PO4:
Molecular Weight of Ca3(PO4)2:
Molecular Weight of H2CO3:
Since the weight of the two reactants are given in the problem, then we need to find out which of them is limiting reagent. Limiting reagent is a substance that gives the least amount of product in the reaction. On the other hand, excess reagent is a substance that gives the excess amount of a substance in the reaction.
Weight of Ca3(PO4)2 from CaCO3:
Weight of Ca3(PO4)2 from H3PO4:
By comparing the weights of Ca3(PO4)2, CaCO3 will give us the least amount of product.
(a) Therefore, the limiting reagent is CaCO3 and the amount of Ca3(PO4)2 produced is
Since CaCO3 is the limiting reagent, then the amount of H3PO4 can be used in the reaction is
(b) Therefore, the excess amount of H3PO4 after the reaction is
