Final answer:
If 2 L of oxygen gas has been formed at STP, it means that 1 mole of oxygen gas has been produced. Since the stoichiometric mole ratio between hydrogen peroxide and oxygen gas is 2:1, the amount of hydrogen peroxide remaining after the formation of 2 L of oxygen gas at STP would be 4 moles.
Step-by-step explanation:
The given equation for the decomposition of hydrogen peroxide is:
2 H₂O₂(aq) → 2 H₂O(l) + O₂(g)
From the balanced equation, we can see that for every 2 moles of hydrogen peroxide, 1 mole of oxygen gas is produced. Therefore, if 2 L of oxygen gas has been formed at STP, it means that 1 mole of oxygen gas has been produced.
Since the stoichiometric mole ratio between hydrogen peroxide and oxygen gas is 2:1, we can calculate the amount of hydrogen peroxide that is required to produce 1 mole of oxygen gas.
Let's assume that x moles of hydrogen peroxide have decomposed. According to the stoichiometry, 2 moles of hydrogen peroxide produce 1 mole of oxygen gas. So,
x mol H₂O₂ → 1 mol O₂
Using the given information that we started with 5 moles of hydrogen peroxide, we can use the stoichiometry to find the remaining amount of hydrogen peroxide:
(5 moles of H₂O₂) → (1 mole of O₂)
Therefore, the amount of hydrogen peroxide remaining after the formation of 2 L of oxygen gas at STP would be 5-1 = 4 moles.