Question

З сасоз+2fepo4 → саз(po4)2+Fe2(CO3)3

How many grams of calcium phosphate
are produced from 26 grams of (Iron Ill)
phosphate and 12.1 grams of calcium
carbonate. what is the excess reactant

Answer 1

A. Excess reactant is FePO4

B. mass of Ca3(PO4)2 produced is 12.41g

Step-by-step explanation:

A. FePO4 has molar mass of 150.82 g/mol => 26/150.82 = 0.17 mole

CaCO3 has molar mass of 100.09 g/mol => 12.1/100.09 = 0.12 mole

3CaCO3 + 2FePO4 => Ca3(PO4)2 + Fe2(CO3)3

2 moles of FePO4 will react with 3 moles of CaCO3

Therefore,

0.17 mole of FePO4 will react with 0.17 x (3/2) = 0.255 mole of CaCO3

and

0.12 mole of CaCO3 will react with 0.12 (2/3) = 0.08 mole of FePO4

Therefore the excess reactant will be FePO4

B. 3 moles of CaCO3 will produce 1 mole of Ca3(PO4)2

so 0.12 mole of CaCO3 will produce 0.12 x (1/3) = 0.04 mole

Ca3(PO4)2 has molar mass of 310.18 g/mol

=> mass of Ca3(PO4)2 produced is 0.04 x 310.18 = 12.41g