Question #cf135

1 Answer
Jun 29, 2015

Answer:

There are actually two ways of approaching this problem.

Explanation:

You're dealing with a redox reaction in which ammonia, #NH_3#, is being oxidized to nitric oxide, #NO#, according to the balanced chemica equation

#color(red)(4)NH_(3(g)) + 5O_(2(g)) -> 4NO_((g)) + 6H_2O_((g))#

Take a look at the #color(red)(1):1# (#color(red)(4):4#) mole ratio that exists between ammonia and nitric oxide. This tells you that regardless of how many moles of ammonia react, the reaction will produce an equal number of moles of nitric oxide.

Keep this in mind, it will become very useful in a short while.

Now, you know that an 8.5-g sample of ammonia reacts to produce 4.5 g of nitric oxide. Before doing anything else, check to see whether or not the aforementioned mole ratio is respected.

Use ammonia and nitric oxide's molar masses to determine how many moles of each reacted/were produced

#8.5cancel("g") * ("1 mole "NH_3)/(17.03cancel("g")) = "0.500 moles"# #NH_3#

and

#4.5cancel("g") * ("1 mole "NO)/(30.01cancel("g")) = "0.150 moles"# #NO#

The number of moles of ammonia and nitric oxide are not equal. Your reaction produced less moles of nitric oxide than the number of moles of ammonia that reacted, which means that one of two things happened

Let's take the first scenario. If not all ammonia reacted, then oxygen must have acted as a limiting reagent. If you work backwards from the number of moles of nitric oxide produced, you have

#0.150cancel("moles"NO) * (color(red)(1)" mole "NH_3)/(1cancel("mole"NO)) = "0.150 moles"# #NH_3#

This means that, out of the 8.5 g sample, only

#0.150cancel("moles") * ("17.03 g" NH_3)/(1cancel("mole")) = "4.5 g"#

of ammonia actually reacted.

In the second scenario, you assume that all the ammonia reacted, but that the reaction doesn't have a 100% yield.

If all the ammonia actually reacted, then the reaction should have produced

#0.450cancel("moles"NH_3) * ("1 mole "NO)/(color(red)(1)cancel("mole"NH_3)) = "0.450 moles"# #NO#

The mass of nitric oxide should have been

#0.450cancel("moles") * ("30.01 g "NO)/(1cancel("mole")) = "13.5 g"# #NO#

This is the reaction's theoretical yield. The actual yield is 4.5 g of nitric oxide produced. This means that the percent yield was

#"% yield" = "actual yield"/"theoretical yield" * 100#

#"% yield" = (4.5cancel("g"))/(13.5cancel("g")) * 100 = "33.3%"#