The reaction of Phosphorous Pentaoxide with water yield Phosphoric Acid as shown below,
P₄O₁₀ + 6 H₂O → 4 H₃PO₄
According to balance equation,
283.88 g (1 mole) P₄O₁₀ requires = 108 g (6 mole) of H₂O
So,
100 g P₄O₁₀ will require = X g of H₂O
Solving for X,
X = (100 g × 108 g) ÷ 283.88 g
X = 38.04 g of H₂O
So, 100 g P₄O₁₀ requires 38.04 g of H₂O, while we are provided with 200 g of H₂O which means that water is in excess and P₄O₁₀ is limiting reagent. Therefore, P₄O₁₀ will control the yield of H₃PO₄. So,
As,
283.88 g (1 mole) P₄O₁₀ produced = 391.96 g (4 mole) of H₃PO₄
So,
100 g P₄O₁₀ will produce = X g of H₃PO₄
Solving for X,
X = (100 g × 391.96 g) ÷ 283.88 g
X = 138.07 g of H₃PO₄
Result:
Theoretical Yield of this reaction is 138.07 g.