Considering the limiting reactant,what is the volume of NO gas produced from 30.0 L of nitrogen gas and 20.0 L of oxygen gas? (Assume constant conditions. ) N₂(g)+ O₂(g) 2 NO(g)

Question

Considering the limiting reactant,what is the volume of NO gas produced from 30.0 L of nitrogen gas and 20.0 L of oxygen gas? (Assume constant conditions. ) N₂(g)+ O₂(g)  2 NO(g)

Quizplus · Accepted Answer

To determine the limiting reactant, we need to first convert the volumes of nitrogen gas and oxygen gas to moles using the ideal gas law. 
n(N₂) = PV/RT = (1 atm)(30.0 L)/(0.08206 L·atm/K·mol)(273 K) = 1.26 mol N₂ 
n(O₂) = PV/RT = (1 atm)(20.0 L)/(0.08206 L·atm/K·mol)(273 K) = 0.84 mol O₂ 
We then use the stoichiometry of the balanced equation to determine the amount of NO gas produced from each of the reactants: 
From N₂: 1.26 mol N₂ × (2 mol NO/1 mol N₂) = 2.52 mol NO 
From O₂: 0.84 mol O₂ × (2 mol NO/1 mol O₂) = 1.68 mol NO 
Since oxygen gas produces less NO gas than nitrogen gas, it is the limiting reactant. Therefore, we use the amount of NO produced from O₂ to determine the volume: 
V(NO) = n(O₂) × (2 mol NO/1 mol O₂) × (22.4 L/mol) = 37.6 L 
However, we need to consider the fact that oxygen gas takes up 20.0 L of the initial mixture, so the maximum volume of NO gas that can be produced is only 20.0 L. Therefore, the correct answer is (C) 40.0 L.

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