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Meehan98

  • one year ago

Need some help with a Chemistry question!

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  1. Meehan98
    • one year ago
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    1 mole of NH3 is 34.08 gram. Do I just convert that back into moles?

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  2. anonymous
    • one year ago
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    2mole of NH3 is 34.08grams

  3. Meehan98
    • one year ago
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    So, isn't it 2 moles of ammonia?

  4. anonymous
    • one year ago
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    Yes, NH3 is ammonia.

  5. Meehan98
    • one year ago
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    How do I figure out how many moles of NH3 are produced in the reaction?

  6. anonymous
    • one year ago
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    You know limiting reactant?

  7. Meehan98
    • one year ago
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    Yes, the limiting reactant is Nitrogen.

  8. Meehan98
    • one year ago
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    No, I'm sorry it's hydrogen.

  9. anonymous
    • one year ago
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    Then use hydrogen and find the amount of ammonia it can be produced. Ignore nitrogen then.

  10. Meehan98
    • one year ago
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    It can produce 102.24 grams of ammonia.

  11. anonymous
    • one year ago
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    \[\frac{ mol hydrogen }{ 1}\times \frac{ mol ammonia }{ mol hydrogen }\]

  12. anonymous
    • one year ago
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    Dont need grams. My formula is a shorter way.

  13. Meehan98
    • one year ago
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    I did nitrogen again..It would be 56.8 grams of ammonia.

  14. Meehan98
    • one year ago
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    Okay, so it can produce 3.3 moles of NH3

  15. anonymous
    • one year ago
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    5 mol H2 x (2mol NH3/3mol H2)

  16. anonymous
    • one year ago
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    Yea.

  17. anonymous
    • one year ago
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    Makes sense, since 3mol H2 produces 2 mol NH3

  18. Meehan98
    • one year ago
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    Thank you! I was making things more difficult than what was needed!

  19. anonymous
    • one year ago
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    5 mol H2 would produce around 3.3 mol NH3

  20. anonymous
    • one year ago
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    No problem!

  21. Meehan98
    • one year ago
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    Quick question: How come we used the limiting reactant (H) instead of the excess reactant (N)?

  22. Photon336
    • one year ago
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    @Meehan98 you always use the limiting reactant to calculate the number of moles and eventually grams of product formed for one simple reason: the limiting reactant runs out first because there is less of it than your reactant that's in excess. once your limiting reactant runs out you cant form any more product because it runs out.

  23. Meehan98
    • one year ago
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    Okay, thank you!! That makes sense.

  24. Photon336
    • one year ago
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    No problem, anytime.. it's just that when one reactant runs out first, think of it this way say we have a reaction in which A+B = c FROM this reaction we need one molecule of A and one of B. to form C. \[A + B --> C \] Say if we had 10 molecules of A and only 5 of B. B would be our limiting reactant because we only have 5 molecules of B and from our reaction we need 1 molecule of A and one molecule of B. so because of this B runs out first, we use the number of moles of B to calculate A.

  25. Photon336
    • one year ago
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    sorry for that last part we use the number of moles of B and then multiply by the ratio of C/B to get how much C is produced.

  26. Meehan98
    • one year ago
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    Okay, what if it's asking how many grams will be produced instead of moles?

  27. Photon336
    • one year ago
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    So from what we said, you need to identify which reactant runs out first and that's going to be our limiting reactant. then you multiply the number of moles of the limiting reactant by the molar ratio to get your product in moles then you can easily say since we have moles of product we multiply that by the molar mass like this: \[moles*\frac{ (molar mass) g }{ moles }\] = grams of product

  28. Photon336
    • one year ago
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    is this clearer?

  29. Meehan98
    • one year ago
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    Yes, it is. Thank you! I'm just having trouble figuring out which is the limited reactant in an equation.

  30. Photon336
    • one year ago
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    here is a problem for you completely made up though A molar mass = 10g/mol B molar mass = 5g/mol 20 grams of A react with 10 grams of B to form C. A + 2B --> C 1. find me the limiting reagent 2. and how many grams of C are produced.

  31. Photon336
    • one year ago
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    FYI this is an important step We have 20 grams of A and 10 grams of B. We must first convert everything to moles. 20grams of A * (1mol/10g) = 2 moles of A 10 grams of B*(1mol/5g) = 2 moles of B

  32. Photon336
    • one year ago
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    So this tells us that we have 2 moles of A and two moles of B. we must find out how many moles we NEED for each. to do this we multiply each by the molar ratio. we look at the formula: we have 1 mole of A and 2 moles of B in he equation. so for A we do \[2 A = \frac{ 2B }{ A } = 4 B\] when we do B \[2 B * \frac{ A }{ 2B } = 1 mol of A\] do you notice something? the way I set this

  33. Photon336
    • one year ago
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    We need 4 moles of B and 1 mol of A but how many moles of each do we have?

  34. Meehan98
    • one year ago
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    Ok, so I think that I use a much more drawn out way of producing the answer than what is needed, but I got B as the limiting reagent and 10 moles of C.

  35. Photon336
    • one year ago
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    yes B is the limiting reagent

  36. Photon336
    • one year ago
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    good, how did you get the number of moles of C?

  37. Meehan98
    • one year ago
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    But, I still need to convert the moles into grams so that will be 10 moles multiplied by the molar mass (5) = 50?

  38. Meehan98
    • one year ago
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    I got the number of moles of C by multiplying 5 mol B times the ratio of C/B which is 10.

  39. Photon336
    • one year ago
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    oh lol sorry in my problem i didn't give you the molar mass of C the imaginary compound; let's say the molar mass of C is 5 grams per mole. show me like by writing out the equation of how you get the grams of C

  40. Meehan98
    • one year ago
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    Based on your explanation above of how to convert from moles to grams of the product, \[\frac{ 5 mol B }{ 1 } X \frac{ mol C }{2 mol B }\] so, I have 10 mol C and you said to multiply that by the molar mass, does that mean by molar mass of C? If so, then I get 50, but I'm still in moles. I'm sorry- not quite grasping this yet.

  41. Photon336
    • one year ago
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    |dw:1441916098745:dw| well first we have identified the limiting reagent which is B. now we multiply by the molar ratio to find the number of moles of C

  42. Photon336
    • one year ago
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    Does this make sense?

  43. Meehan98
    • one year ago
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    Yes, I was using 5 mol of B when I multiplied the ratio instead of 2.

  44. Photon336
    • one year ago
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    okay so put your information together like this LIMITING REAGENT B 2 moles of B Molar ratio of B/C C/2B why did I put B on the bottom?

  45. Meehan98
    • one year ago
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    I'm not sure, why?

  46. Meehan98
    • one year ago
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    Is it because it's the limiting reagent? because I would've written it like: 2molB/1molC

  47. Photon336
    • one year ago
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    B was our limiting reagent

  48. Photon336
    • one year ago
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    and we had 2 moles of B

  49. Photon336
    • one year ago
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    Next we must figure out the number of moles of C

  50. Photon336
    • one year ago
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    \[2 mol B * \frac{ C }{ 2B }\] = 1 mol C

  51. Photon336
    • one year ago
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    I want you to look at tis last sentence. when we multiplied by the molar ratio we were looking for C, right? so when we set it up we make sure that B is in the denominator and you can see that B cancels out leaving us with C? make sense?

  52. Meehan98
    • one year ago
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    Yes, I know how to do that; I just didn't understand now that we have 1 mol C, how to convert to grams like the question asked.

  53. Photon336
    • one year ago
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    \[1 mol * (\frac{ 5 grams }{ mol } ) = \]

  54. Photon336
    • one year ago
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    set it up like this general rule \[moles*(\frac{ molar mass grams }{ mol })\]

  55. Meehan98
    • one year ago
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    So, C would be 5 grams.

  56. Photon336
    • one year ago
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    exactly

  57. Meehan98
    • one year ago
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    Okay, a lot to take in, but helped immensely!! Thank you for spending time and explaining everything out! I truly appreciate it!

  58. Photon336
    • one year ago
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    no problem

  59. Photon336
    • one year ago
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    look up dimensional analysis; it will help with these problems

  60. Meehan98
    • one year ago
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    Ok, will do:)

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