Q90P - Chemistry: Molecular Nature Of Matter And Change

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?`

Step-by-Step Solution

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 1: 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.

Step 2: 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 = ${\mathbf{O}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{13}}{\mathbf{2}}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{055}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3575}\mathbf{mole}$  

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

Step 3: Moles of H 2 O 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.

Step 4: 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 =  $\mathbf{0}\mathbf{.}\mathbf{495}\mathbf{\times }\mathbf{6}\mathbf{.}\mathbf{023}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{molecules}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{98}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{molecules}$