Chemistry

I posted this question earlier, but I believe it was a bit confusing in the way I typed it.

Given the following two step reaction:

4 P + 5 O2 -> P4O10

P4O10 + 6 H2O -> 4H3PO4

If you have 8.28 g of P, 15.65 g of O2, and 19.95 g of H2O, how many grams of H3PO4 can be produced?
--------------------------
I've converted everything to moles:
4P:
8.28g
m=g/mw
x=8.28/124.0
x=0.067

502:
15.65g
x=15.65g/160
x= 0.0978

6H2O:
19.95g
x=19.96/108.0
0.1848

However, now I am stuck.
Do I need to find the limited?
If so, how do I go about this?


Please and Thank You

  1. 👍 0
  2. 👎 0
  3. 👁 110
asked by Bradley
  1. And it still is confusing. Yes, you do need to go through a limiting reagent. First, the molar mass of P (atomic mass) is 30.97 BUT P actually is P4 so let's re-write the equation as
    P4 + 5O2 --> P4O10
    moles P = 8.28/123.895 = 0.06683 mols P.
    moles O = 15.65/32 = 0.4891 moles oxygen.
    We have two scenarios depending upon which is the limiting reagent. I prefer to do it this way; i.e., determine the moles P4O10 produced under each.
    a. If we have 0.06683 moles P4 and all the oxygen we need, how much P4O10 will we get?
    0.06683 x (1 mole P4O10/1 mole P4) = 0.06683 x (1/1) = 0.06683 moles P4O10.

    b. If we have 0.4891 moles oxygen and all the P4 we need.
    0.4891 x (1 mole P4O10/5 moles O2) = 0.09781 moles P4O10.

    c. The smaller number is ALWAYS the correct answer in limiting reagent problems; therefore, the limiting reagent is P4 and we have 0.06683 reacting and producing 0.06683 moles P4O10.

    Now you are ready to tackle the second part of the reaction. It's another limiting reagent problem. Just follow the steps above.

    1. 👍 0
    2. 👎 0
    posted by DrBob222

Respond to this Question

First Name

Your Response

Similar Questions

  1. RE: Math Urgent

    I posted this earlier but I think I phrased it in a confusing way Find g(x) + g(x) if g(x) = {(-4,5), (-2,-3), (0,5), (1,3), (3,-1), (5,5)} and g(x) = {(-3,-2), (-2,6), (-1,0), (1,0), (2,-6), (3,5), (4,0), (6,-2)}. I don't need

    asked by Anonymous on January 21, 2013
  2. Math

    Whoever can help with a question I posted earlier. I am very sorry; I put the wrong variable in the question. Question is to solve for y and not "x" as my previous message and the equation is: x = 3(r + ty) So sorry for the

    asked by Cassandra on April 18, 2015
  3. advance functions

    will someone please help me with the question i posted earlier ? :(

    asked by oranges on May 25, 2011
  4. algebra 1

    Reiny I posted a question earlier (about 3:00) and I have a question about your solution. How did you get -4x? I came up with -8x which does not allow for anymore reduction. Please let me know how you did it. Thanks!

    asked by Dippy on March 9, 2009
  5. damon can you help with this?

    I posted this question earlier but had a typo, can you take another look? { 1/3x + 3/4 y - 2/3 z = -8 x + 1/2 y - 1/3z = 18 1/6 x - 1/8 y - z = -24 a. (-6, 8, -24) b. (-6, -8, -24) c. (0, 8, -24) d. (6, 8, 24) e. (6, -8, 24)

    asked by lottie on August 31, 2016
  6. Physics

    This question was posted earlier but the explanation given was confusing. The three 500 g masses in the daigram are connected by massless, rigid rods to form a triangle. What is the triangle's rotational energy (in J) if it

    asked by Me on November 14, 2011
  7. Math

    Sorry posted the wrong question earlier. Let A and B be sets. Show that A \ (A \ B) ⊆ B.

    asked by Winston on January 27, 2017
  8. math

    for the question i posted earlier, pie = 3. im sorry if i confused some of you people. sorry.

    asked by brianna on December 14, 2008
  9. Why can't I post

    I don't know why, but i am getting a friend to post this up for me, but i cannot post a question, when i do, the successful window never comes up. it just goes back to the forum and doesnt add. can anyone help me why? or what i

    asked by becky on January 18, 2007
  10. Physics - please check over

    Could I please get some further assistance on a question I posted earlier? I'm not getting the correct answer and I'm not sure what I'm doing wrong.

    asked by Susan on April 15, 2009

More Similar Questions