For the following reaction, 27.0 grams of diphosphorus pentoxide are allowed to react with 8.95 grams of water. diphosphorus pentoxide (s) + water (l) phosphoric acid (aq) What is the maximum amount of phosphoric acid that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

0.331 moles of H₃PO₄

Limiting reagent: H₂O

Amount of the excess reagent that remains: 3.41 g of P₂O₅

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

the increase of protons   hope this

16

explanation:

the marks go by 4s