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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