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6 Nov 2019

One of the steps in the commercial process for converting ammoniato nitric acid is the conversion of NH3to NO:

4NH3(g)+5O2(g)?4NO(g)+6H2O(g)

In a certain experiment, 1.25gof NH3reacts with 2.29gof O2.


Which is the limiting reactant?

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Irving Heathcote
Irving HeathcoteLv2
22 Sep 2019

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Related questions

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water.

What is the maximum mass of H2O that can be produced by combining 68.3 g of each reactant?

4NH3 (g) + 5O2 (g) -----> 4NO (g) + 6H2O (g)

The first step in the Ostwald process for producing nitric acid is as follows:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g).

If the reaction of 15.0 g of ammonia with 15.0 g of oxygen gas yields 8.70 g of nitric oxide, what is the percent yield of this reaction?

29.0%

32.9%

49.5%

61.8%

77.3%

The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps:

4NH3(g)+5O2(g)------->4NO(g)+6H2O(g) -908 kj/mol (delta H)

2NO(g)+O2(g)----------->2NO2(g) -112 kj/mol

3NO2(g)+H2O(l)-------->2HNO3(aq)+NO(g) -140 kj/mol

Write the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations (Water is also a product).

So i know the answer is this...

"Multiply the equations so that the coefficients of the N-products become the same as the N-educts of the following line.
4NH3(g)+5O2(g)------->4NO(g)+6H2O(g) -908 kj/mol (delta H) (x3)
2NO(g)+O2(g)----------->2NO2(g) -112 kj/mol (x 6)
3NO2(g)+H2O(l)-------->2HNO3(aq)+NO(g) -140 kj/mol (x 4)

12NH3(g)+15O2(g)------->12NO(g)+18H2O(...
12NO(g)+6O2(g)----------->12NO2(g)
12NO2(g)+4H2O(l)-------->8HNO3(aq)+4NO.."

What i want to know is why do we multiply the equations like that? Am i missing a concept? Because why can we just not add the eqautions together?

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