Refer to the following equation: 4NH3(g)+7O2(g)=4NO2(g)+6H2O(g). How many molecu

Question

Refer to the following equation: 4NH3(g)+7O2(g)=4NO2(g)+6H2O(g). How many molecules of NO2 are produced when 5.38 mil of ammonia is completely reacted? Refer to the following equation: 4NH3(g)+7O2(g)=4NO2(g)+6H2O(g). How many molecules of NO2 are produced when 5.38 mil of ammonia is completely reacted? Refer to the following equation: 4NH3(g)+7O2(g)=4NO2(g)+6H2O(g). How many molecules of NO2 are produced when 5.38 mil of ammonia is completely reacted? Refer to the following equation: 4NH3(g)+7O2(g)=4NO2(g)+6H2O(g). How many molecules of NO2 are produced when 5.38 mil of ammonia is completely reacted?

Explanation / Answer

Given : mol NH3 = 5.38 mol

Reaction :

4 NH3 + 7 O2 ------------ > 4 NO2 (g) + 6 H2O g)

Here we use stoichiometric equivalence

4 mol NH3 produces 3 mol NO2

We have to calculate moles of NO3 produced from 5.38 moles of NH3

Conversion factor is : (4 mol NO2 / 4 mol NH3 )

Lets use this to get moles of NO2 from 5.38 moles of NH3

= 5.38 mol NH3 * 4 mol NO2 / 4 mol NH3

= 5.38 mol NO2

Ans: 5.38 mol NO2 are produced

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