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 limiting reagent.
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
