Question

50g of CaCO3 is allowed to react with 73.5g of H3PO4 calculate amount of Ca3(po4)2 formed in moles

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

Answer 2

o.667 is it's answer.. .........