How many moles of oxygen are consumed when 100.L of dinitrogen pentoxide are produced in the following equation?
2 N₂(g) + 5 O₂(g) $\rarr$ 2 N₂O₅(g)

Question

How many moles of oxygen are consumed when 100.L of dinitrogen pentoxide are produced in the following equation? 2 N₂(g)+ 5 O₂(g)$\rarr$ 2 N₂O₅(g)

Quizplus · Accepted Answer

Given the balanced chemical equation, 2 N2O5(g) + 5 O2(g) → 2 N2O5(g), it's clear that 5 moles of O2 are consumed to produce 2 moles of N2O5. Since 100 L of N2O5 are produced and using the ideal gas law (assuming standard conditions where 1 mole = 22.4 L), the moles of N2O5 produced are 100 L / 22.4 L/mol = 4.46 moles. Therefore, to find the moles of O2 consumed, we use the stoichiometric ratio from the equation, which is 5 moles of O2 for every 2 moles of N2O5 produced. Thus, (5/2) * 4.46 moles = 11.15 moles of O2, which rounds to 11.2 moles.

How Many Moles of Oxygen Are Consumed When 100 →\rarr→ 2 N2O5(g) A)0

How Many Moles of Oxygen Are Consumed When 100 $\rarr$ 2 N₂O₅(g)
A)0