If 7.0 mol of no and 5.0 mol of o2 are reacted tegethor. the reaction generates 3.0 mol of no2. what is the percent yield for the reaction? 2no(g) + o2(g) -> 2no2(g)

The answer to your question is: 43 %

Explanation:

Data

NO = 7 mol

O2 = 5 mol

NO2 = 3 mol

percent yield = ?

Reaction

                             2NO(g)    +   O2(g)  ⇒    2NO2(g)      

Proportion of reactants

From the reaction             2 moles of NO / 1 moles of O2 = 2

From the experiment        7 moles of NO / 5 moles of O2 = 1.4

Then, the limiting reactant is NO.

Rule of three

                            2 moles of NO 2 moles of NO2

                            7 moles of NO   x

                            x = (7 x 2) / 2

                            x = 7 moles of NO2

% yield = experimental/ theoretical x 100

% yield = 3/7 x 100

% yield = 42.9 ≈ 43                                                    

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