Question

In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2 g of water are obtained. Determine the percent yield of the reaction.

Answer 1

The percent yield of the reaction : 89.14%

Further explanation

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):

`(80)/(17)=4.706`

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

`\tt (120)/(32)=3.75`

Mol ratio of reactants(to find limiting reatants) :

`\tt (4.706)/(4)/ (3.75)/(5)=1.1765/ 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)`

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

`\tt (6)/(5)* 3.75=4.5`

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

`\tt \%yield=(actual)/(theoretical)* 100\%\\\\\%yield=(72.2)/(81)* 100\%=89.14\%`