# Balance the following reaction: CaCO3 + FePO4 –> Ca3(PO4)2 + Fe2(CO3)3. Which reactant is limiting, assuming we start with 100 grams of calcium carbonate and 45 grams of iron (III) phosphate. What is the mass of each product that can be formed? What m…

##3CaCO_3##+ ##2FePO_4## –> ##Ca_3(PO_4)2## + ##Fe_2(CO_3)3## (a)
As per the above equation (a) Three moles of ##CaCO_3## , consumes two moles of ##FePO_4##.
In terms of mass one mole of ##CaCO_3## , has mass 100 g/mol.
and one mole of ##FePO_4## has mass 150.8 g/mol.
so let us set up the ratio;
##(“3 mole” CaCO_3)/(“2 mole” FePO_4)## = ##(300g)/(301.6g)## (b)
X g of ##CaCO_3## will consume 45 g of ##FePO_4## (c)
equating two equation (b) and (c)
##(300g)/(301.6g)## = ##”Xg”/”45g”##
300 x 45 = 301.6 X
301.6 X = 13500
X = ##13500/301.6## = 44.7 g => 45 g
So, 45 g of ##CaCO_3## will react with 45 g of ##FePO_4## . The amount of ##FePO_4## added is 45 g. The amount of ##CaCO_3## remains unused is 100-45= 55 g. Iron (III) phosphate is a .
2 moles of ##FePO_4## produces 1 mole of ##Ca_3(PO_4)2##
301.6 g of ##FePO_4## produces 310.17 g ##Ca_3(PO_4)2## (d)
45 ##FePO_4## produces X g ##Ca_3(PO_4)2## (e)
##”301.6g”/”310.17g”## = ##”45g”/”Xg”##
301.6 ( X ) = 45 x 310.17
301.6 (X) = 13957.65
X = ##13957.65/301.6## = 46.27 g