Question

Butane gas is compressed and used as a liquid fuel in disposable cigarette lighters and lightweight camping stoves. Suppose a lighter contains 5.50 mL of butane (d =0.579 g/mL).

(a) How many grams of oxygen are needed to burn the butane completely?

(b) How many moles of H₂O form when all the butane burns?

(c) How many total molecules of gas form when the butane burns completely?`

Short answer

(a) Mass of ${\mathbf{O}}_{\mathbf{2}}$reacted = $\mathbf{32}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{3575}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{44}\mathbf{g}$

(b) Moles of${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$formed are 0.275 mole.

(c) Total molecules of gas formed = $\mathbf{2}\mathbf{.}\mathbf{98}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{molecules}$

Step-by-step solution

Writing balanced equation

First of all, let us check the balanced equation:

$\mathbf{2}{\mathbf{C}}_{\mathbf{4}}{\mathbf{H}}_{\mathbf{10}}\left(g\right)\mathbf{+}\mathbf{130}\left(g\right)\mathbf{\to }\mathbf{8}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{10}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)$

Now we can calculate the asked questions one by one.

Grams of oxygen needed

Mass of butane used

$\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{butane}\mathbf{=}\mathbf{volume}\mathbf{\times }\mathbf{density}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{50}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{579}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{1845}\mathbf{g}$

Moles of butane is:

$\mathbf{Mole}\mathbf{}\mathbf{of}\mathbf{}\mathbf{butane}\mathbf{=}\frac{\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{butane}}{\mathbf{Molar}\mathbf{}\mathbf{Mass}}\mathbf{=}\frac{\mathbf{3}\mathbf{.}\mathbf{1845}}{\mathbf{58}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{055}\mathbf{mole}$

From the balanced equation we can say that 2 moles of butane react with 13 moles of${\mathbf{O}}_{\mathbf{2}}$

Now, 0.055 moleof butane will react with= role="math" localid="1657103490605" ${\mathbf{O}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{13}}{\mathbf{2}}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{055}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3575}\mathbf{mole}$

Mass of reacted = $\mathbf{32}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{3575}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{44}\mathbf{g}$

Moles of H2O  formed

From the balanced equation it is clear that 2 moles of butane gives 10 moles of${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$so 0.055 mole of butane will give

=$\frac{\mathbf{10}}{\mathbf{2}}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{055}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{275}\mathbf{mole}$

So, moles of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ formed are 0.275 mole.

Total gaseous molecules formed  

From the balanced equation it is clear that 2 moles of butane give 18 moles of gases (i.e.,${\mathbf{CO}}_{\mathbf{2}}$and${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$), so 0.055 mole of butane will give

=$\frac{\mathbf{18}}{\mathbf{2}}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{055}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{495}\mathbf{mole}$

1 mole of a gas =$\mathbf{6}\mathbf{.}\mathbf{023}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{molecules}$

So, 0.495 mole of gases = $\mathbf{0}\mathbf{.}\mathbf{495}\mathbf{\times }\mathbf{6}\mathbf{.}\mathbf{023}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{molecules}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{98}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{molecules}$