Question

question :

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50g of caco3 is allowed to react with 70g of h3po4 .

calculate:

1. amount of ca3(po4)2 produced.
2. amount of unreacted reagent.

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Answer 1

The balanced chemical equation is:

3 CaCO3 + 2 H3PO4 = Ca3(PO4)2 + 3 CO2 + 3 H2O

Molar mass of:

CaCO3 = 100

H3PO4 = 98

Ca3(PO4)2 = 310

Moles of:

CaCO3 = 50/100 = 0.5

H3PO4 = 73.5/98 = 0.75

From the equation, 3 moles of CaCO3 react with 2 moles of H3PO4

0.5 moles of CaCO3 will react with = (0.5x2)/3 = 0.333 moles of H3PO4

But we have 0.75 moles of H3PO4

CaCO3 is the limiting reactant

From the equation, 3 moles of CaCo3 produce 1 mole of Ca3(PO4)2  

0.5 moles of CaCO3 will produce = (0.5x1)/3 = 0.1667 moles of Ca3(PO4)2  

0.1667 moles of Ca3(PO4)2  will be produced