The balanced equation for the reaction is
CO(g) + 2H₂(g) ⇄ CH₃<span>OH(g)
Since given concentrations are at equilibrium state, the expression for the equilibrium constant, k can be written as
k = [</span>CH₃OH(g)] / [CO(g)] [H₂(g) ]²
By substitution,
k = 0.030 M / 0.020 M x (<span>0.072 M</span>)²
k = 289.35 M⁻²
Answer:
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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):
mass O₂ = 120 g
mol O₂(MW=32 g/mol) :
Mol ratio of reactants(to find limiting reatants) :
mol of H₂O based on O₂ as limiting reactants :
mol H₂O :
mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :
The correct option is A.
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