Chemistry Forum

I am doing a project at the minute with three reactions occurring. The reactions include:

CH₄ + H₂O --> CO + 3H₂... Step 1
CO + H₂₀ <--> CO₂ + H₂... Step 2
CH₄ + 2H₂O <--> CO₂ + 4H₂... Step 3

My lecturer and I have assumed the first reaction has a CH₄ conversion of 95% with the flow rate of both the CH₄ and H₂O being 134.4753 mol/s. If I have the 95% conversion, I am wondering how much H₂O would be provided to each of the other two steps; that is assuming the other 5% of the CH₄ is going to the other steps.

First, Step 2 doesnt consume methane, so the yield could be everything of this one, called x%.
Second, Step 3 is the summery of step 1 and 2, so it would be (95% * x%)/100%.
In practise it has to be checked how much CO₂ in comparison of H₂ is obtained.