Answer:
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