50 g of CaCO3 is allowed to react with 73.5 g of H3PO4. Calculate:(i)

1.	50 g of CaCO3 is allowed to react with 73.5 g of H3PO4. Calculate:(i) Amount of Cas(PO4)2 formed (in moles) Ca (03 -t H(ii)Amount of unreacted reagent (in moles)
Answer» molecular mass of CaCO3 = 100gm so, here number of moles are = 50/100 = 0.5 molecular mass of H3PO4 is = 98 so, moles = 73.5/98 = 0.75 now , balanced equation is 3CaCO3+2H3PO4=Ca3(PO4)2+3H2CO3 and since 0.5/3 is smaller that 0.75/2 , so CaCO3 is the limiting reagent now amount of Ca3(PO4)2 , will be = 0.5/3 = 0.1666.. moles and the amount of H3PO4 reacted is 2*0.5/3 = 0.3333. at the end , all CaCO3 will react so, moles of H3PO4 left will be 0.75-0.33 = 0.42

50 g of CaCO3 is allowed to react with 73.5 g of H3PO4. Calculate:(i) Amount of Cas(PO4)2 formed (in moles) Ca (03 -t H(ii)Amount of unreacted reagent (in moles)

Answer»

molecular mass of CaCO3 = 100gm so, here number of moles are = 50/100 = 0.5

molecular mass of H3PO4 is = 98 so, moles = 73.5/98 = 0.75

now , balanced equation is 3CaCO3+2H3PO4=Ca3(PO4)2+3H2CO3

and since 0.5/3 is smaller that 0.75/2 , so CaCO3 is the limiting reagent

now amount of Ca3(PO4)2 , will be = 0.5/3 = 0.1666.. moles

and the amount of H3PO4 reacted is 2*0.5/3 = 0.3333.

at the end , all CaCO3 will react

so, moles of H3PO4 left will be 0.75-0.33 = 0.42