Given this equation:

2P2O5 + 6H2O ---> 4H3PO4.

If you begin with 4.8 grams of p2o5 and 15.2 grams h20, what will be your limiting reactant.

I got the answer to that and it is P2O5. (6.6 grams) and H2O was 55.2 grams. so the limiting reactant P2O5.

Then the next question is how many grams of the excess reagent remain unreacted?

That is what i need help with.. can anyone help me? :)

And the answer is 13.4.

You are correct that P2O5 is the limiting reagent.

You already have the moles P2O5 from your previous calculation. Convert that to moles H2O.
Convert moles H2O to grams.
Subtract from the original 15.2 to determine the amount H2O remaining.
Post your work if you get stuck. 13.4 g H2O remaining is the right answer.

how do i know what the excess reagent is?

Thanks for your help! :)

I'm stuck...
so i did 4.8 g P2O5 * 1mol/142. I got .03. then what do i do after that?

I think i figured it out... can you tell me if this is right.

i did 4.8 g P2O5 * 1mol/142 *.03 mols P2O5/ 1g * 6H20 / 2P2O5*18.0G/1. Instead of getting 13.4, I got 13.58. Is that still right. If not can you please show me how you did it! Thanks again, I really appreciate it:)

You know the OTHER chemical given to you (in this case it's water) is the one in excess.

4.8 g P2O5 *(1 mol P2O5/142) = 0.0338 mols.
mols H2O needed to react with this much P2O5 = 0.0338 mols P2O5 x (6 mols H2O/2 mols P2O5) = 0.101 mols H2O.

grams H2O = mols x molar mass = 0.101 mols x (18 g/1 mol) = 1.82 grams.

Amount to start = 15.2 g.
Amount used by P2O5 = 1.82
15.2 - 1.82 = 13.37 which rounds to 13.4 g.

To determine the amount of excess reagent remaining, we first need to identify the limiting reactant, which in this case is P2O5. This means that all of the H2O will be consumed in the reaction, while P2O5 will not be completely used up.

To find the amount of excess reagent remaining, we need to calculate the amount of P2O5 that reacts based on the stoichiometry of the balanced equation.

The molar ratio between P2O5 and H3PO4 is 2:4, or simplified to 1:2. This means that for every 2 moles of P2O5, 4 moles of H3PO4 are produced.

First, convert the mass of P2O5 to moles using its molar mass:
4.8 g P2O5 / (141.95 g/mol) = 0.0338 mol P2O5

Since the ratio is 1:2, this means that 0.0338 mol P2O5 will produce 0.0338 x 2 = 0.0676 mol of H3PO4.

Next, convert the moles of H3PO4 to grams using its molar mass:
0.0676 mol H3PO4 x (97.99 g/mol) = 6.633 grams of H3PO4

Since we initially had 15.2 grams of H2O, which is the excess reagent, and all of it reacts, we subtract the mass of H3PO4 formed from the initial mass of H2O:
15.2 g H2O - 6.633 g H3PO4 = 8.567 grams of H2O remaining unreacted

So, the amount of excess reagent, H2O, remaining unreacted is 8.567 grams.