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Lena772

  • one year ago

If at the beginning of the reaction you used 4.30 g NH3 and 15.76 g O2, what is the theoretical yield (in grams) of NO (you will need to find the limiting reagent and then the theoretical yield of NO from this reagent)?

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  1. Lena772
    • one year ago
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    \[4NH _{3} + 5O _{2} →4NO +6H _{2} O\]

  2. aaronq
    • one year ago
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    The limiting reagent is found by dividing the moles of the reactants by their respective stoichiometric coefficients (from the balanced reaction they're participating in). Compare these values, the lowest is the limiting reagent.

  3. Lena772
    • one year ago
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    4.30g NH3 * 1 mol / 17.03g NH3 = 0.252 mol NH3 15.76g 02 * 1 mol / 32 g O2 = 0.4925 mol O2

  4. Lena772
    • one year ago
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    0.252/4=0.063 0.4925/5 = 0.0985 O2 would be the limiting reagent

  5. aaronq
    • one year ago
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    nope, 0.063<0.0985

  6. Lena772
    • one year ago
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    okay... so now what do I do?

  7. aaronq
    • one year ago
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    Use the moles from NH3 to find the theoretical yield. it's the exact same process as the last question

  8. Lena772
    • one year ago
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    0.063*17.03= 1.07289 g

  9. Lena772
    • one year ago
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    1.07g ?

  10. aaronq
    • one year ago
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    yes

  11. aaronq
    • one year ago
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    wait the molar mass of NO is not 17.03 g/mol

  12. aaronq
    • one year ago
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    it's 30.01 g/mol

  13. Lena772
    • one year ago
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    right. So it would be 1.89 then ?

  14. aaronq
    • one year ago
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    yep

  15. aaronq
    • one year ago
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    everything seems to be in order

  16. Lena772
    • one year ago
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    Were the coefficients for the reactants supposed to be used anywhere? @aaronq

  17. aaronq
    • one year ago
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    yeah, when you compare NH3 to NO, you need to use the coefficients, but since both are 4, it doesn't make a difference

  18. Lena772
    • one year ago
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    Hmm I don't know why I'm getting it wrong then

  19. Lena772
    • one year ago
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    You don't use the coefficient in the initial conversion of grams to moles @aaronq ?

  20. aaronq
    • one year ago
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    nope, mass to moles interconversions only involve the molar mass \(\sf moles=\dfrac{mass}{Molar~mass}\)

  21. aaronq
    • one year ago
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    idk either, you're using the correct number of sig figs too

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