For this question we have to start with balanced chemical equation.

##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