Question

under certain conditions the reaction of ammonia with excess oxygen will produce a 29.5 percent yield of NO.what mass of nh3 must react with excess oxygen to yield 157g NO

Answer 1

4NH3 (g) + 5O2 (g) --> 4 MO (g) + 6H2O (g)

To use this reaction equation you should convert the 157 g of NO into moles NO:

moles NO = 157 gNO/(30 g/molNO) = 5.23 moles NO

According to the problem statement, this amount of NO should represent a 29.5% yield, so a 100% (perfect) yield should be given by:

100% yield = 5.23 moles NO/0.295 = 17.7 moles NO

Now we can use the reaction equation to determine how much NH3 we need. According to the equation we need 4 moles of NH3 for every 4 moles of NO potentially produced(see the above reaction). this is just one for one so we need the same 17.7 moles of NH3.

convert this back into weight of NH3:

gNH3 = 17.7 molNH3 x 17g/gmol NH3 = 300.9 gNH3