Question

under certain conditions the reaction of ammonia with excess oxygen will produce a 29.5% yield of NO. What mass of NH3 must react with excess Oxygen to yield 157g of NO?

4NH3+5O2=4NO+6H20

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

Answer 2

Step1 Moles of NO= 157/30

Step2 Moles of NH3= Moles of NO= 157/30

Step3 % Yield of NO is 29.5; ; Moles of NH3 = 157x100/30x29.5

Step4 Mass of NH3 = Moles of NH3x17 = 157x100x17/30x29.5 = 301.58 g

Answer 3

Hi!

Our first step is to calculate the target moles of NO to be produced using the given mass in grams and the molar mass of NO:

157g NO × (1 mole NO / 30 grams) = 5.23 moles NO

Next, we see from the problem that following the reaction formula, we will only get a 29.5% yield. So, we'll need to account for this.

If we were to get a 100% yield, we would need the following:

5.23 moles NO × (4 moles Ammonia / 4 moles NO) = 5.23 moles Ammonia

But, since we are only getting 29.5% yield, we will perform the calculation like this:

5.23 moles NO × (4 moles Ammonia / (0.295·4 moles NO)) = 17.74 moles Ammonia

Finally, we convert the required moles of ammonia to grams using its molar mass:

17.74 moles Ammonia × (17 grams / 1 mole) = 301.58 grams Ammonia

I hope this helps! Please don't forget to rate : )