Lena772
  • Lena772
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)?
Chemistry
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jamiebookeater
  • jamiebookeater
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Lena772
  • Lena772
\[4NH _{3} + 5O _{2} →4NO +6H _{2} O\]
aaronq
  • aaronq
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.
Lena772
  • Lena772
4.30g NH3 * 1 mol / 17.03g NH3 = 0.252 mol NH3 15.76g 02 * 1 mol / 32 g O2 = 0.4925 mol O2

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Lena772
  • Lena772
0.252/4=0.063 0.4925/5 = 0.0985 O2 would be the limiting reagent
aaronq
  • aaronq
nope, 0.063<0.0985
Lena772
  • Lena772
okay... so now what do I do?
aaronq
  • aaronq
Use the moles from NH3 to find the theoretical yield. it's the exact same process as the last question
Lena772
  • Lena772
0.063*17.03= 1.07289 g
Lena772
  • Lena772
1.07g ?
aaronq
  • aaronq
yes
aaronq
  • aaronq
wait the molar mass of NO is not 17.03 g/mol
aaronq
  • aaronq
it's 30.01 g/mol
Lena772
  • Lena772
right. So it would be 1.89 then ?
aaronq
  • aaronq
yep
aaronq
  • aaronq
everything seems to be in order
Lena772
  • Lena772
Were the coefficients for the reactants supposed to be used anywhere? @aaronq
aaronq
  • aaronq
yeah, when you compare NH3 to NO, you need to use the coefficients, but since both are 4, it doesn't make a difference
Lena772
  • Lena772
Hmm I don't know why I'm getting it wrong then
Lena772
  • Lena772
You don't use the coefficient in the initial conversion of grams to moles @aaronq ?
aaronq
  • aaronq
nope, mass to moles interconversions only involve the molar mass \(\sf moles=\dfrac{mass}{Molar~mass}\)
aaronq
  • aaronq
idk either, you're using the correct number of sig figs too

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