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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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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\[4NH _{3} + 5O _{2} →4NO +6H _{2} O\]
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.
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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0.252/4=0.063 0.4925/5 = 0.0985 O2 would be the limiting reagent
nope, 0.063<0.0985
okay... so now what do I do?
Use the moles from NH3 to find the theoretical yield. it's the exact same process as the last question
0.063*17.03= 1.07289 g
1.07g ?
yes
wait the molar mass of NO is not 17.03 g/mol
it's 30.01 g/mol
right. So it would be 1.89 then ?
yep
everything seems to be in order
Were the coefficients for the reactants supposed to be used anywhere? @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
Hmm I don't know why I'm getting it wrong then
You don't use the coefficient in the initial conversion of grams to moles @aaronq ?
nope, mass to moles interconversions only involve the molar mass \(\sf moles=\dfrac{mass}{Molar~mass}\)
idk either, you're using the correct number of sig figs too

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