Question

Help!! How many liters of methane gas (CH4) need to be combusted to produce 12.4 liters of water vapor, if all measurements are taken at the same temperature and pressure? Show all of the work used to solve this problem.

CH4 (g) + 2 O2 (g) yields CO2 (g) + 2 H2O (g)

Answer 1

@aaronq @zepdrix Stoichiometry help?

Answer 2

Do you have a general idea of how to do this or are you starting from scratch?

Answer 3

I'll say scratch.

Answer 4

This is the general scheme for any stoichiometry problem, i can work it through with you after you've read this.

Stoichiometry

When performing stoichiometric calculations the values to be used \(\sf \color{red}{must~be~in~moles}\)*. This is because the Molar Masses (\(\sf M_{m}\) of two elements are not the same. This is to say that 1 gram of Hydrogen atoms is not the same number of atoms as 1 gram of Nitrogen atoms. We need to be working with numbers of particles (e.g. atoms, molecules, formula units) and using moles is the way this is done.

*A shortcut can be taken when working with \(\sf \color{blue}{Ideal~Gases}\) as moles are \(\propto\) volume. And, so volume can be regarded as moles, where 22.414 L = 1 mol.

General Scheme:
\(\sf \large 1.\)First write and balance an equation for the process described.
\(\sf \large 2.\)Next, use the stoichiometric coefficients to find moles produced.
\(\sf \large 3.\)Set up a ratio using the species of interest, like so (for a general reaction):

\(\sf \large \color{red}{a}A + \color{blue}{b}B \rightleftharpoons \color{green}{c}C\)

here upper case letters (A,B,C) are the chemical species, and lower case letters (a,b,c) are the coefficients,

\(\sf \dfrac{moles_A}{\color{red}{a}}=\dfrac{moles_B}{\color{blue}{b}}=\dfrac{moles_C}{\color{green}{c}}\)

You can use any pair from here, depending on what the question is asking for. For example: if you have 2 moles of B, how many moles of C can you produce? Now rearrange algebraically,

\(\sf\dfrac{2}{\color{blue}{b}}=\dfrac{moles _C}{\color{green}{c}}\rightarrow moles _C=\dfrac{2*\color{green}{c}}{\color{blue}{b}}\)


\(\sf \large 4.\) Solve

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Answer 5

You needn't know much to do this...
Anyway, just take a look at the equation. It says that one mole of methane gives two moles of water vapor. According to Gay-Lussac's law of Gaseous Volumes, when gases react, they do so in volumes that bear a simple ratio with one another and with the products so formed, provided that all the measurements pertaining to the volumes are made with respect to the same temperature and pressure. Here, the RATIO is the ratio of the number of moles, which is 1 : 2(in case of methane and water vapor). Therefore,
1 : 2 = x : 12.4 (where x liters is the volume of methane)
Solving for x, you get 6.2 liters. So simple as it is...

Answer 6

while you are 100% correct, kizhakke, I wanted to show the OP how to do any stoichiometry problem- not just this one.