Answer:
64.8L of H2
Step-by-step explanation:
We need to start with a balanced chemical equation. I arrive at the following:
O2 + 2H2 = 2H2O
This tells us that we'll get 2 moles of H2O for every 1 mole O2 consumed, and that we'll need 2 moles of H2 for every 1 mole of O2. For this question, the important ratio is 2 moles of H2 for every 1 mole of O2. We know we have 23.4 L of O2. Let's convert that into moles O2 by using a very useful conversion factor that works for ALL gases at STP (Standard Temperature and Pressure). The conversion is 22.4M/L. That means that at STP we'll have 1 mole of that gas if we have 22.4L of the gas. And that works for all gases. [1mole = 22.4L at SDTP for all gases].
So let's calculate the moles of oxygen gas present in 32.4L:
(32.4L)*(1 mole/22.4L)= 1.45 moles O2.
If all the oxygen reacts, we'll need twice the number of moles of H2, as per the balanced equation.
(1.45 moles O2)*(2) = 2.90 moles H2
Convert 2 moles H2 at STP into moles H2: (2.90 moles H2)*(22.4L/mole) = 64.8L of H2.