Question

uppose the reaction Ca3(PO4)2 + 3H2SO4 ï‚® 3CaSO4 + 2H3PO4 is carried out starting with 153 g of Ca3(PO4)2 and 87.6 g of H2SO4. How much phosphoric acid will be produced?

Answer 1

The answer is 58.4 g of H₃PO₄

Given that mass of Ca₃(PO₄)₂ is 153g

mass of H₂SO₄ is 87.6g

We need to calculate the mass of H₃PO₄

So the balanced chemical reaction is

Ca₃(PO₄)₂  +  3H₂SO₄  ⇒   3CaSO₄  +  2H₃PO₄

Let us 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

Now let us calculate the limiting reactant

The Theoretical Yield =  Ca₃(PO₄)₂ / H₂SO₄

= 310 / 3(98)

= 1.05

The Experimental yield

Ca₃(PO₄)₂ / H₂SO₄

= 153 / 87.6 = 1.74

Because the observed percentage was more than the predicted proportion, H2SO4 is the limiting reactant.

Let us Calculate the molar mass of H₃PO₄

H₃PO₄ = (1 x 3) + (31 x 1) + (16 x 4)

= 3 + 31 + 64

= 98 g

Now Calculate the mass of H₃PO₄

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

87.6 g of H₂SO₄ ------------------  x

x = ( 87.6 x 2 x 98) / (3 x 98)

x = 17169.6 / 294

x = 58.4g of H₃PO₄

Learn more about Theoretical Yield here

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