GIVEN: 4NH3 + 5O2===> 4NO + 6H2O

Now, in a container 6.0moles of NH3 are mixed with 8.0moles of O2. If the reaction produces 8.0moles of H2O, what is the percent yield of this reaction?

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To find the percent yield of a reaction, you need to compare the actual amount of the desired product produced to the theoretical amount of the product that should have been produced based on stoichiometry.

Step 1: Determine the limiting reagent. This is the reactant that will be completely consumed and determines the maximum amount of product that can be formed. To do this, compare the moles of NH3 and O2 present with their stoichiometric coefficients.

NH3: 6.0 moles
O2: 8.0 moles

From the balanced equation, you can see that the stoichiometric ratio between NH3 and O2 is 4:5. Therefore, we need to convert the moles of NH3 and O2 to a common unit to compare them.

NH3: 6.0 moles x (5 moles O2 / 4 moles NH3) = 7.5 moles O2
O2: 8.0 moles

Since there is an excess of O2 (8.0 moles > 7.5 moles), NH3 is the limiting reagent.

Step 2: Calculate the theoretical yield. This is the maximum amount of product that can be formed based on the limiting reagent.

From the balanced equation, the stoichiometric ratio between NH3 and H2O is 6:6 (or 1:1 because they have the same coefficient).

NH3: 6.0 moles
H2O: ? moles

Since NH3 is the limiting reagent, the moles of H2O formed will be equal to the moles of NH3. Therefore, the theoretical yield of H2O is 6.0 moles.

Step 3: Calculate the percent yield. This is the ratio of the actual yield (given as 8.0 moles of H2O) to the theoretical yield, multiplied by 100.

Percent yield = (Actual yield / Theoretical yield) x 100
Percent yield = (8.0 moles / 6.0 moles) x 100
Percent yield = 133.33%

Therefore, the percent yield of this reaction is 133.33%.