Question

How many moles of methane are produced when 25.1 moles of carbon dioxide gas react with excess hydrogen gas?

Answer 1

Hint: Carbon dioxide gas can be converted into methane gas upon addition of hydrogen gas. This reaction is accompanied by loss of water molecules. The balanced equation of the reaction can help us determine the moles of products formed from given moles of reactants.
\[C{{O}_{2}}(g)+4{{H}_{2}}(g)\xrightarrow{{}}C{{H}_{4}}(g)+2{{H}_{2}}O(g)\]

Complete answer:
To solve this question, we first need to understand the concept of mole ratio.
A conversion factor which relates the number of moles of any two components of a chemical reaction is known as the mole ratio. The coefficients in a balanced chemical equation of a reaction are used to determine the numbers in a conversion factor.
We can see that in the balanced chemical equation, one mole of carbon dioxide $C{{O}_{2}}$ gas reacts with four moles of hydrogen gas ${{H}_{2}}$ to form one mole of methane $C{{H}_{4}}$ and two moles of water ${{H}_{2}}O$.
Now we have to calculate the number of moles of methane formed when 25.1 moles of carbon dioxide gas, in presence of excess hydrogen.
To do so, we have to find out the mole ratio between carbon dioxide and methane in the given reaction. Since it is given to us that there is an excess of hydrogen, we need not be concerned about any mole ratio involving hydrogen.
Since, one mole of carbon dioxide gas forms one mole of methane gas, the mole ratio is
\[\dfrac{\text{1 mol C}{{\text{H}}_{4}}}{\text{1 mol C}{{\text{O}}_{2}}}\]
Now multiplying the mole ratio with the given number of moles of carbon dioxide gas, we get
\[\begin{align}
  & =\dfrac{\text{1 mol C}{{\text{H}}_{4}}}{\text{1 mol C}{{\text{O}}_{2}}}\times 25.1\text{ mol C}{{\text{O}}_{\begin{smallmatrix}
 2 \\
\end{smallmatrix}}} \\
 & =25.\text{1 mol C}{{\text{H}}_{4}} \\
\end{align}\]
Hence it can be seen that when 25.1 moles of carbon dioxide gas react with excess hydrogen gas, 25.1 moles of methane gas are produced.

Note:
One must note that different relative amounts of reactants and products can also be considered by using the same proportionality. For example, if the coefficients of each of the compound is multiplied by 10 molecules, then ten molecules of carbon dioxide $C{{O}_{2}}$ gas reacts with forty molecules of hydrogen gas ${{H}_{2}}$ to form ten molecules of methane $C{{H}_{4}}$ and twenty molecules of water ${{H}_{2}}O$.