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.

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  1. 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.

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  2. how do i know what the excess reagent is?

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  3. 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?

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  4. 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:)

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  5. 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.

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