The percent yield of the reaction : 89.14%
Reaction of Ammonia and Oxygen in a lab :
4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)
mass NH₃ = 80 g
mol NH₃ (MW=17 g/mol):
[tex]\dfrac{80}{17}=4.706[/tex]
mass O₂ = 120 g
mol O₂(MW=32 g/mol) :
[tex]\tt \dfrac{120}{32}=3.75[/tex]
Mol ratio of reactants(to find limiting reatants) :
[tex]\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)[/tex]
mol of H₂O based on O₂ as limiting reactants :
mol H₂O :
[tex]\tt \dfrac{6}{5}\times 3.75=4.5[/tex]
mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :
[tex]\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%[/tex]
