Question

calculate the number of moles of oxygen gas that are required to react with 4.74 mol dodecane, C 10H 22. 2 C 12H 26(l) + 37 O 2(g) → 24 CO 2(g) + 26 H 2O(g)?

Answer 1

Approximately
`87.7\; \rm mol`.

Step-by-step explanation:

Dodecane is the name for an alkane. There are twelve carbon atoms in each dodecane molecule. Its molecular formula is
`\rm C_(12)H_(26)`. (In an alkane molecule with
`n` carbon atoms, there would be
`2\, n + 2` hydrogen atoms.)

As seen in the question, the balanced equation for this reaction is:

`\rm 2\; C_(12)H_(26)\, (l) + 37\; O_2\, (g) \to 24\; CO_2\, (g) + 26\; H_2O\, (g)`.

The question gives the value of
`n\left(\mathrm{C_(12)H_(26)}\right)` (the number of moles of dodecane in the reaction) and is asking for
`n\left(\mathrm{O}_2\right)` (the number of moles of oxygen in this reaction.) It would be helpful if there is a ratio
`\displaystyle \frac{n\left(\mathrm{O_2}\right)}{n\left(\mathrm{C_(12)H_(26)}\right)}` for directly converting

Such a ratio does exist. Best of all, it can be found directly from the coefficients of the balanced chemical equation. In the balanced equation for this reaction,

• the coefficient of
  `\mathrm{C_(12)H_(26)}` is
  `2`, and
• the coefficient of
  `\mathrm{O}_2` is
  `37`.

In other words, for each mole of the reaction that took place,
`2\; \rm mol` of
`\mathrm{C_(12)H_(26)}` would react with
`37\;\rm mol`. Note that in this example, there's nothing special about the quantity of one mole. It is possible to scale the quantities of both reactants, and this ratio would still hold for this reaction. Thus,

`\begin{aligned}&\frac{n\left(\mathrm{O_2}\right)}{n\left(\mathrm{C_(12)H_(26)}\right)} \\ &= \frac{\text{Coefficient of $\mathrm{O_2}$ in balanced chemical equation}}{\text{Coefficient of $\mathrm{C_(12)H_(26)}$ in balanced chemical equation}} \\ &= (37)/(2)\end{aligned}`.

Apply this ratio to find
`n\left(\mathrm{O}_2\right)`:

`\begin{aligned}& n\left(\mathrm{O_2}\right)\\ &= \frac{n\left(\mathrm{O_2}\right)}{n\left(\mathrm{C_(12)H_(26)}\right)} \cdot n\left(\mathrm{C_(12)H_(26)}\right) \\ &= (37)/(2) \,n\left(\mathrm{C_(12)H_(26)}\right) \approx 87.7\; \text{mol} \end{aligned}`.