Question

Suppose the reaction Ca3(PO4)2 + 3H2SO4 --> 3CaSO4 + 2H3PO4 is carried out starting with

153 g of Ca3(PO4)2
and 76.8 g of H2SO4
How many grams of phosphoric acid will be produced.​

Answer 1

The answer to your question is 51.2 g of H₃PO₄

Step-by-step explanation:

Data

mass of Ca₃(PO₄)₂ = 153 g

mass of H₂SO₄ = 76.8 g

mass of H₃PO₄ = ?

Balanced chemical reaction

Ca₃(PO₄)₂ + 3H₂SO₄ ⇒ 3CaSO₄ + 2H₃PO₄

Process

1.- Calculate the molar mass of the reactants

Ca₃(PO₄)₂ = (3 x 40) + (2 x 31) + (8 x 16) = 120 + 62 + 128 = 310 g

H₂SO₄ = (1 x 2) + (32 x 1) + (16 x 4) = 2 + 32 + 64 = 98 g

2.- Calculate the limiting reactant

Theoretical yield Ca₃(PO₄)₂ / H₂SO₄ = 310 / 3(98) = 1.05

Experimental yield Ca₃(PO₄)₂ / H₂SO₄ = 153 / 76.8 = 1.99

The limiting reactant is H₂SO₄ because the experimental proportion was higher than the theoretical proportion.

3.- Calculate the molar mass of H₃PO₄

H₃PO₄ = (1 x 3) + (31 x 1) + (16 x 4) = 3 + 31 + 64 = 98 g

4.- Calculate the mass of H₃PO₄

3(98) g of H₂SO₄ ------------------ 2(98) g of H₃PO₄

76.8 g of H₂SO₄ ------------------ x

x = (76.8 x 2 x 98) / (3 x 98)

x = 15052.8 / 294

x = 51.2 g of H₃PO₄