1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Limiting Reactant Problems Stoichiometry

  1. Mar 14, 2006 #1
    Please help me I am so lost!!

    Here's the lesson. My writing is in Blue.

    Where should I start? I am that confused; I feel like :cry:

    Thanks for your help!!
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 14, 2006 #2
    first of all you need to convrt everything in grams to MOLES

    this is the reaction that is going to take place
    [tex] S +O_{2} \rightarrow SO_{2} [/tex]

    so if u had 1 mol of S and 2 mol of O2, how many moles of SO2 would u form??
    use the same concept to your problem, but convert to moles first!
  4. Mar 15, 2006 #3
    Be sure to find the limiting reactant first, before calculating the mass of sulfur dioxide for each respective quantity of reactants.
  5. Mar 15, 2006 #4
    To find the limiting reagent, its somewhat or rather like your blood question, u need to find the number of moles of the substances that are involved.

    So, for your Question 1, convert the mass given to the SI units.Remember to always write the correct equation out and BALANCE it or else you will get the wrong mole ratio!!

    Equation: S (s) + O2 (g) -> SO2(g)

    Now, from the periodic table, you would have found that the MR of the S & O2 to be 32.1 and (16.0 x 2) = 32.0 respectively.
    No. of moles in S = mass of S / Mr of S
    = 20.0 / 32.1
    = 0.6231 mol ( to 4 sig. fig.)

    From this value, and usin the ratio that was given in the eqn above, we can deduce that for COMPLETE REACTION of the S will require 0.6231 mole of the O2 as well!!

    No. of moles in O2 = mass of O2/ Mr of O2
    = 160 / 32.0
    = 5.00 mol
    Since from the values, we can see that the O2 is present in excess and hence, the S is the limiting reagent!!

    So, since the mole ratio is 1:1:1 respectively in the eqn,
    this imply that the no of moles in the SO2 is also 0.6231 mol!!

    Findin Mr of the SO2 = 32.1 + 16.0 x 2 = 64.1
    Hence, mass of SO2 = 64.1 x 0.6231 = 39.9 g ( to 3 sig. fig.)

    Try qns 2 on your own now!!=)
  6. Mar 17, 2006 #5

    S + O2 = SO2
    32.066 + 31.9988 = 64.0648
    1 mole + 1 mole = 2 moles?

    Is that it? Wouldn't it only form one mole of SO2?

    Thanks for replying!
  7. Mar 17, 2006 #6
    Please help me a little bit more! Thanks so much!!
  8. Apr 29, 2007 #7
    can you guys just post the answer
  9. Apr 29, 2007 #8
    im verry confused. what does all this mean

    No. of moles in O2 = mass of O2/ Mr of O2
    = 160 / 32.0
    = 5.00 mol
    Since from the values, we can see that the O2 is present in excess and hence, the S is the limiting reagent!!

    So, since the mole ratio is 1:1:1 respectively in the eqn,
    this imply that the no of moles in the SO2 is also 0.6231 mol!!
  10. Apr 29, 2007 #9


    User Avatar

    Staff: Mentor

    Why? You could as well calculate mass of the product for every reactant - and then select the lowest number.

  11. Apr 29, 2007 #10


    User Avatar

    Staff: Mentor

  12. Apr 26, 2008 #11
    okay i really need help with this...
    How many grams SO2 can be formed from 20.0 g of S and 160g O2?
    my equation is S + O --> SO2
    and i change grams to moles or something like that??
    someone pleaseee help me figure this out! im so confused...
  13. Apr 27, 2008 #12
    you have to work in moles. Calculate how many moles of Sulphur and Oxygen you have.

    S + O2 ---> SO2

    Now, the stoichiometric equation says, 1 mol S reacts with 1 mol O2 to give 1 mol SO2.

    say from the number of moles you calculated you got 0.5 mol O2 and 0.75 mol S. (an example)

    from this you find that S is in larger amount than O2. Since 1 mol O2 reacts with 1 mol S, the 0.5 mol O2 will completely react with 0.5 mol of S. therefore 0.25 mol of S remains. the limiting reagent is O2.

    also, according to the equation, 1 mol of O2 produces 1 mol of SO2.
    but you have 0.5 mol of O2 which reacted. this means that 0.5 mol SO2 will be produced.

    now you find the mass of the 0.5 mol SO2.
  14. Apr 27, 2008 #13
    okayy i get it!!!
    so for this particular problem the one with the lower number of moles is the limting reagant??
  15. Apr 27, 2008 #14


    User Avatar

    Staff: Mentor

    Lower number of moles is not enough - you have to account for stoichiometric coefficients. In this case all stoichiometric coefficients are 1, so just looking at number of moles is enough.

    pH meter
  16. Apr 27, 2008 #15
    what are stoichiometric coefficients

    and why do i have to account for them as well??
    im confused... :confused:
  17. Apr 27, 2008 #16
    say you had H2SO4 + 2NaOH -----> Na2SO4 + 2H2O

    you have 0.3 mol H2SO4 and 0.4 mol NaOH.

    which one is limiting?
  18. Apr 28, 2008 #17
    the limiting one is H2SO4..right?
  19. Apr 28, 2008 #18
    bah no.....

    that's why you have to look at the molar coefficients of each reactant. according to the equation, 2 mol of NaOH will react with 1 mol of H2SO4.

    but you have 0.4 mol NaOH. this amount will react with 0.2 mol H2SO4. All the NaOH will be over and 0.1 mol of H2SO4 will remain. hence the NaOH is limiting.

    try this one,

    2H3PO4 + 3Ba(OH)2 -----> Ba3(PO4)2 + 6H2O

    say you have 0.5 mol H3PO4 (phosphoric acid) and 0.9 mol Ba(OH)2 (Barium Hydroxide)

    which one is limiting?
  20. Apr 28, 2008 #19
    because the number infront of H3PO4 is 2...right??
  21. Apr 28, 2008 #20
    you determine it like this:

    from equation: 2 mol H3PO4 reacts with 3 mol Ba(OH)2
    we have 0.5 mol H3PO4 which will therefore react with 0.75 mol Ba(OH)2.

    this shows that Ba(OH)2 is in excess, only 0.75 mol will be used up out of the 0.9 mol present. Phosphoric acid is limiting.

    if you had done it the other way round...

    3 mol Ba(OH)2 reacts with 2 mol H3PO4.

    we have 0.9 mol Ba(OH)2 which would therefore react with 0.6 mol H3PO4.
    but we only have 0.5 mol H3PO4. hence H3PO4 is limiting, all the Ba(OH)2 will not react.

  22. Apr 28, 2008 #21
    ohhhh i get it!!
    i was thinking of it like backwards.
    thankks so muchh
  23. Apr 28, 2008 #22
    does anyone know how to do enthalpys??
    here is the problem i have to do... (its in red)

    Assignment Problem
    Balance the equation and calculate the enthalpy change for the following reaction:

    NH3(g) + O2(g) --> N2(g) + H2O(l)

    NH3(g)= -46.11 kj/mol

    O2(g) = 0 kJ/mol

    N2 = 0 kJ/mol

    H20= - 285.830 kJ/mol

    i think the balanced equation is 4NH3 + 3O2 --> 2N2 +6H2O, but im not sure how to do the enthalpy change.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook