natasha.mi3015 wrote:Hi, I am from a non-science background and at the beginning of my studies so I am probably a bit slow picking up on "in between" steps in the explanation to a solution; I apologise in advance but I am finding it difficult to follow the explanation of the solution.

No apologies necessary, you should be congratulated for your efforts!

natasha.mi3015 wrote:- How do you find that 3 moles of C + 6 moles of H2 produce 3 moles of CH4?

The passage provides the balanced reaction:

(iii) Creation of Methane:

2H2 + C → CH4

If we can figure out the number of moles of H2 and C, then stoichiometry (the balanced reaction) allows us to conclude the number of moles methane CH4. Notice that the number of moles in the reaction (iii) above can be represented as: 2:1:1 (meaning H2:C:CH4).

The Explanation shows how weight is converted to the number of moles. Some students like to memorize that the number of moles n = (weight)/(molecular weight), this can also be represented (dimensional analysis) as: moles = (grams)/(grams/mole) which shows that 'grams' cancel leaving the answer in moles.

From the Explanation, we know that there are 6 moles of H2 and 3 moles of CO2. But CO2 is not in equation (iii) so we need to find a relationship between CO2 and C. The preceding is found in equation (ii) where the relationship between the 2 molecules can be represented as 1:1, so in this case, 3:3. Now we know that 6 moles of H2 reacts with 3 moles of C. Since the ratio of (iii) is represented as 2:1:1 (meaning H2:C:CH4), multiplying through be 3 we get: 6:3:3. So 6 H2 react with 3 C to make 3 CH4.

Since: moles = (grams)/(grams/mole)

Then: moles(grams/mole) = (grams) (notice how 'moles' cancels)

So for methane: 3 (16) = 48

You may want to watch (or re-watch!) the video Stoichiometry for Chapter 1 (also found in the Videos section of gamsat-prep.com). Yes, it does get easier as you complete more practice questions!

Please feel open to asking further questions to clarify what you encounter.