Question

1. One of the common fertilizer used to enrich soil with phosphate is

superphosphate, CaH4(PO4)2. Typically, superphosphate is produced by reacting some form of phosphate rock with sulfuric acid. In one preparation, calcium phosphate is reacted with sulfuric acid according to the reaction Ca3(PO4)2 +2H2SO4  CaH4(PO4)2 + 2CaSO4

20,000 kg/day of raw calcium phosphate containing 14wt% inert impurities is reacted with 15,000 kg/day of 92wt% H2SO4 in water. 95% of the limiting reactant is converted. The molecular weight of Ca3(PO4)2 and H2SO4 are 310 and 98, respectively
a) What is the limiting reactant? Show your work
b) What are the flow rates (in kmol/day) of the exiting species

Answer 1

Balanced chemical reaction

Ca3(PO4)2 +2H2SO4 = CaH4(PO4)2 + 2CaSO4

Mass of Ca3(PO4)2 = 20000*0.86 = 17200 kg/day

Mass of H2SO4 = 15000*0.92 = 13800 kg/day

From the stoichiometry of the reaction

1 kmol of Ca3(PO4)2 reacts with = 2 kmol of H2SO4

310 kg of Ca3(PO4)2 reacts with = 2*98 = 196 kg of H2SO4

17200 kg of Ca3(PO4)2 reacts with = 196*17200/310

=10874.83 kg of H2SO4

But we have 13800 kg H2SO4 which is greater than required 10874.83 kg

Excess reagent = H2SO4

Limiting reagent = Ca3(PO4)2

Part b

Conversion of Ca3(PO4)2 = 0.95*17200

= 16340 kg/day / 310 kg/kmol

= 52.709 kmol/day

Conversion of H2SO4 = 2*52.709 = 105.42 kmol/day

At the exit

Molar flow rate of CaH4(PO4)2 = 52.709 kmol/day

Molar flow rate of CaSO4 = 2*52.709 = 105.42 kmol/day

Unconverted Ca3(PO4)2 in the product = initial - converted

= 17200/310 - 52.709

= 2.775 kmol/day

Unconverted H2SO4 in the product = initial - converted

= 13800/98 - 105.42

= 35.396 kmol/day

Water = (15000 kg/day*0.08) /(18kg/kmol)

= 66.666 kmol/day