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rewona [7]
3 years ago
8

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.
Chemistry
1 answer:
valentinak56 [21]3 years ago
6 0

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

\dfrac{80}{17}=4.706

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

\tt \dfrac{120}{32}=3.75

Mol ratio of reactants(to find limiting reatants) :

\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

\tt \dfrac{6}{5}\times 3.75=4.5

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%

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(a) pH = 12.73

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