Considering the reaction stoichiometry and the definition of limiting reagent, the percent yield for the reaction is 42.86%.
The balanced reaction is:
2 NO + O₂ → 2 NO₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) and use a simple rule of three as follows: if by stoichiometry 1 mole of O₂ reacts with 2 moles of NO, if 5 moles of O₂ react how much moles of NO will be needed?
[tex]moles of NO=\frac{5 moles of O_{2} x2 moles of NO}{1 mole of O_{2} }[/tex]
moles of NO= 10 moles
But 10 moles of NO are not available, 7 moles are available. Since you have less moles than you need to react with 5 moles of O₂, NO will be the limiting reagent.
Then you can apply the following rule of three, considering the limiting reagent: if by stoichiometry 2 moles of NO produce 2 moles of NO₂, 7 moles of NO will produce how many moles of NO₂?
[tex]amount of moles of NO_{2} =\frac{7 moles of NOx2 moles of NO_{2}}{2 moles of NO}[/tex]
amount of moles of NO₂= 7 moles
So, the amount of moles of NO₂ produced is 7 moles.
The maximum amount of product that can be produced is called the theoretical yield.
Percent yield is the amount of product actually obtained from a reaction and determines the efficiency of the reaction.
The expression for percent yield is:
[tex]Percent yield=\frac{actual yield}{theoretical yield}x100[/tex]
To determine the percent yield for the reaction in this case, you know:
Replacing in the definition of percent yield:
[tex]Percent yield=\frac{3 moles}{7 moles}x100[/tex]
Percent yield= 42.86%
Finally, the percent yield for the reaction is 42.86%.
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