Considering the limiting reactant,what is the volume of NO gas produced from 50.0 L of ammonia gas and 60.0 L of oxygen gas? (Assume constant conditions. ) 4 NH₃(g)+ 5 O₂(g) 4 NO(g)+ 6 H₂O(g)

Question

Considering the limiting reactant,what is the volume of NO gas produced from 50.0 L of ammonia gas and 60.0 L of oxygen gas? (Assume constant conditions. ) 4 NH₃(g)+ 5 O₂(g)  4 NO(g)+ 6 H₂O(g)

Quizplus · Accepted Answer

To determine the limiting reactant, we need to calculate the amount of product that would be formed if all of each reactant was used. 
For ammonia gas, the amount of NO produced would be: 
(50.0 L NH₃) x (4 mol NO/4 mol NH₃) x (22.4 L/mol) = 50.0 L NO
For oxygen gas, the amount of NO produced would be: 
(60.0 L O₂) x (4 mol NO/5 mol O₂) x (22.4 L/mol) = 53.8 L NO
Therefore, oxygen gas is the limiting reactant and the amount of NO produced is determined by the amount of oxygen gas available. Using the stoichiometry of the balanced chemical equation: 
(60.0 L O₂) x (4 mol NO/5 mol O₂) x (22.4 L/mol) = 53.8 L NO

Considering the Limiting Reactant,what Is the Volume of NO Gas