Question

Suppose the reaction Ca3(PO4)2 + 3H2SO4 - 3CaSO4 + 2H3PO4 is carried out starting with 127 g of Ca3(PO4)2 and 71.9 g of H2SO4. How much phosphoric acid will be produced?

Answer 1

1)

Molar mass of Ca3(PO4)2,

MM = 3*MM(Ca) + 2*MM(P) + 8*MM(O)

= 3*40.08 + 2*30.97 + 8*16.0

= 310.18 g/mol

mass(Ca3(PO4)2)= 127.0 g

use:

number of mol of Ca3(PO4)2,

n = mass of Ca3(PO4)2/molar mass of Ca3(PO4)2

=(1.27*10^2 g)/(3.102*10^2 g/mol)

= 0.4094 mol

Molar mass of H2SO4,

MM = 2*MM(H) + 1*MM(S) + 4*MM(O)

= 2*1.008 + 1*32.07 + 4*16.0

= 98.086 g/mol

mass(H2SO4)= 71.9 g

use:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(71.9 g)/(98.09 g/mol)

= 0.733 mol

Balanced chemical equation is:

Ca3(PO4)2 + 3 H2SO4 ---> 2 H3PO4 + 3 CaSO4

1 mol of Ca3(PO4)2 reacts with 3 mol of H2SO4

for 0.4094 mol of Ca3(PO4)2, 1.228 mol of H2SO4 is required

But we have 0.733 mol of H2SO4

so, H2SO4 is limiting reagent

we will use H2SO4 in further calculation

Molar mass of H3PO4,

MM = 3*MM(H) + 1*MM(P) + 4*MM(O)

= 3*1.008 + 1*30.97 + 4*16.0

= 97.994 g/mol

According to balanced equation

mol of H3PO4 formed = (2/3)* moles of H2SO4

= (2/3)*0.733

= 0.4887 mol

use:

mass of H3PO4 = number of mol * molar mass

= 0.4887*97.99

= 47.89 g

Answer: 47.9 g

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