The first step in the ostwald process for producing nitric acid is 4nh3(g) + 5o2(g) -> 4no(g) + 6h2o(g). if the reaction of 150. g of ammonia with 150. g of oxygen gas yields 87. g of nitric oxide (no), what is the percent yield of this reaction?

The percentage yield of the given reaction is 77.33%.

Explanation:

Moles is calculated by using the formula:

Given mass of ammonia = 150g

Molar mass of ammonia = 17 g/mol

Putting values in above equation, we get:

Given mass of oxygen = 150g

Molar mass of oxygen = 32 g/mol

Putting values in above equation, we get:

For the given chemical equation:

By Stoichiometry,

5 moles of oxygen reacts with 4 moles of ammonia.

So, 4.6875 moles of oxygen will react with = of ammonia

As, moles of ammonia required is less than the calculated moles. Hence, ammonia is present in excess and is termed as excess reagent.

Therefore, oxygen is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the given reaction:

5 moles of oxygen gas produces 4 moles of nitric oxide

So, 4.6875 moles of oxygen gas will produce = of nitric oxide

Now, to calculate the theoretical amount of nitric oxide, we use equation 1:

Molar mass of nitric oxide = 30 g/mol

Given mass of nitric oxide = 112.5 g

Now, to calculate the percentage yield, we use the formula:

Experimental yield = 87 g

Theoretical yield = 112.5 g

Putting values in above equation, we get:

Hence, the percentage yield of the given reaction is 77.33%.

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.

explanation: no but i do not usually comment