Calculate the theoretical yield if 100.0 g p4o10 react with 200.0 g h2o

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.

The correct name for n3-ion should be azide.