Answer:
0.331 moles of H₃PO₄
Limiting reagent: H₂O
Amount of the excess reagent that remains: 3.41 g of P₂O₅
Step-by-step explanation:
Let's discover the reaction where the reactants are diphosphorus pentoxide and water and the product is phosphoric acid
P₂O₅ (s) + 3H₂O(l) → 2H₃PO₄ (aq)
Let's convert the mass of the reactants to moles:
27 g / 141.94 g/mol = 0.190 moles of pentoxide
8.95 g / 18 g/mol = 0.497 moles of water
Ratio is 1:3, 1 mol of pentoxide reacts with 3 moles of water
0.190 moles of pentoxide will react with (0.190 . 3) /1 = 0.570 moles of water
We only have 0.497 moles and we need 0.570, so the limiting reagent is the water.
Ratio is 3:2, 3 moles of water will produce 2 moles of phosphoric acid
Then, 0.497 moles must produce (0.497 . 2) /3 = 0.331 moles of H₃PO₄
Now we know that the excess reagent is the pentoxide.
3 moles of water need 1 mol of pentoxide to react
Therefore 0.497 moles of water will react with (0.497 .1) / 3 = 0.166 moles of pentoxide.
As we have 0.190 moles of pentoxide, we only need 0.166 moles, so
(0.190 -0.166) = 0.024 moles of P₂O₅ still remains after the reaction is complete. Let's convert the moles to mass:
0.024 mol . 141.94 g/ 1 mol = 3.41 g of P₂O₅