# Question #e484d

Mar 21, 2014

$3 C a C {O}_{3}$+ $2 F e P {O}_{4}$ --> $C {a}_{3} \left(P {O}_{4}\right) 2$ + $F {e}_{2} \left(C {O}_{3}\right) 3$ (a)

As per the above equation (a) Three moles of $C a C {O}_{3}$ , consumes two moles of $F e P {O}_{4}$.

In terms of mass one mole of $C a C {O}_{3}$ , has mass 100 g/mol.
and one mole of $F e P {O}_{4}$ has mass 150.8 g/mol.

so let us set up the ratio;

$\left(\text{3 mole" CaCO_3)/("2 mole} F e P {O}_{4}\right)$ = $\frac{300 g}{301.6 g}$ (b)

X g of $C a C {O}_{3}$ will consume 45 g of $F e P {O}_{4}$ (c)

equating two equation (b) and (c)

$\frac{300 g}{301.6 g}$ = $\text{Xg"/"45g}$

300 x 45 = 301.6 X

301.6 X = 13500

X = $\frac{13500}{301.6}$ = 44.7 g => 45 g

So, 45 g of $C a C {O}_{3}$ will react with 45 g of $F e P {O}_{4}$ . The amount of $F e P {O}_{4}$ added is 45 g. The amount of $C a C {O}_{3}$ remains unused is 100-45= 55 g. Iron (III) phosphate is a limiting reagent.

2 moles of $F e P {O}_{4}$ produces 1 mole of $C {a}_{3} \left(P {O}_{4}\right) 2$

301.6 g of $F e P {O}_{4}$ produces 310.17 g $C {a}_{3} \left(P {O}_{4}\right) 2$ (d)

45 $F e P {O}_{4}$ produces X g $C {a}_{3} \left(P {O}_{4}\right) 2$ (e)

$\text{301.6g"/"310.17g}$ = $\text{45g"/"Xg}$

301.6 ( X ) = 45 x 310.17

301.6 (X) = 13957.65

X = $\frac{13957.65}{301.6}$ = 46.27 g