Question

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)?

Answer 1

\[4NH _{3} + 5O _{2} →4NO +6H _{2} O\]

Answer 2

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.

Answer 3

4.30g NH3 * 1 mol / 17.03g NH3 = 0.252 mol NH3

15.76g 02 * 1 mol / 32 g O2 = 0.4925 mol O2

Answer 4

0.252/4=0.063

0.4925/5 = 0.0985

O2 would be the limiting reagent

Answer 5

nope, 0.063<0.0985

Answer 6

okay... so now what do I do?

Answer 7

Use the moles from NH3 to find the theoretical yield. it's the exact same process as the last question

Answer 8

0.063*17.03= 1.07289 g

Answer 9

1.07g ?

Answer 10

yes

Answer 11

wait the molar mass of NO is not 17.03 g/mol

Answer 12

it's 30.01 g/mol

Answer 13

right. So it would be 1.89 then ?

Answer 14

yep

Answer 15

everything seems to be in order

Answer 16

Were the coefficients for the reactants supposed to be used anywhere? @aaronq

Answer 17

yeah, when you compare NH3 to NO, you need to use the coefficients, but since both are 4, it doesn't make a difference

Answer 18

Hmm I don't know why I'm getting it wrong then

Answer 19

You don't use the coefficient in the initial conversion of grams to moles @aaronq ?

Answer 20

nope, mass to moles interconversions only involve the molar mass

\(\sf moles=\dfrac{mass}{Molar~mass}\)

Answer 21

idk either, you're using the correct number of sig figs too