[eixty]

Hi, I need some help with figuring out how to solve this problem.


The question is this:

Under appropriate conditions, the reaction between methane (CH₄) gas and steam proceeds as shown by the equation:
CH₄(g) + 2 H₂O(g) ---> CO₂(g) + 4 H₂(g)

How many liters of each of  the product can be produced from the following volumes of reactants? Assume all gases are at the same condition.

(a) 45.0 L of CH₄ and 45.0 L of H₂O
(b) 45.0 L of CH₄ and 65.0 L of H₂O
(c) 64.0 L of CH₄ and 134.0 L of H₂O

answers:
(a) 22.5 L CO₂, 90 L H₂
(b) 33.0 L CO₂, 132 L H₂
(c) 64.0 L CO₂, 256 L H₂

I tried to do the first problem like this:

Step 1: 45 CH₄ x CO₂/CH₄

Step 2: 45 CH₄ x CO₂/CH₄

Answer for me would be 45 CO₂

The book answer is 22.5 L CO₂. Why is that? At first I thought it was because I hadn't balanced the equation but when I looked at it, it is. Thanks

[enahs]

Lets just start with (A)
When you have 45.0L of methane and 45L of water, look at your equation. To react completely you need twice as much water as methane. You only have equal amount. Now, you can reason out that because I only have half of the reactant quantity of water required, I will only get half of the products.

Or, you can divide your stoichmetric coefficients by whatever it takes to get your limiting reagent to unity (1). That is, if you divide all those coefficients by 2 in this case (because 2/2 will get you 1), you get:
 ½ methane +1 water=1/2 carbon dioxide +2 hydrogen. So you are starting off with the 45L of water as the limiting reagent, and just multiply it by the products levelized stoichmetric coefficients to determine the quantity of product.


(B) Again, water vapor is your limiting reagent, you do not have twice as much of it as.
Again, make the stoichmetric coefficient in front of the limiting reagent 1 (same as in A, and multiply the amount of limiting reagent by the stoichmetric coefficient of the product)


C is the same, but the limiting reagent is now the methane.


Note, the answer for B is wrong, it is 32.5L of methane and 132 L of water.



*edit*
I feel is if that explanation is confusing/not worded very well. If it is I can try and explain it in a different/better way.

[eixty]

½CH4 + 1 H₂O ----> ½CO₂ + 2 H₂O

Thats how it would look for (a) and (b) correct?

(a) 45 x ½CO₂ = 22.5 L CO₂
     45 x  2H₂ = 90.0 L H₂

(b) 65 x ½CO₂ = 32.5 L CO₂
     65 x  2H₂ = 130.0 L H₂

1 CH4 + 2 H₂O ----> 1 CO₂ + 4 H₂O

Since for (c) methane is limiting regent, the formula stays the same since methane is already at a unity of 1?

(c) 64 x 1CO₂ = 64.0 L CO₂
     64 x 4H₂ = 256.0 L H₂


Is that how it is done? Thanks for your help and patience.