Hydrocarbons: Alkanes and Alkenes: Question 8

Worked solution

Part (a): Complete combustion of propane

In a plentiful (“excess”) supply of oxygen, propane burns completely to form carbon dioxide and water.

Start with the unbalanced skeleton equation: $C_3H_8 + O_2 \rightarrow CO_2 + H_2O$

Balance carbon first (3 carbon atoms in propane need 3 $CO_2$): $C_3H_8 + O_2 \rightarrow 3CO_2 + H_2O$

Balance hydrogen next (8 hydrogen atoms in propane need 4 $H_2O$): $C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O$

Now count oxygen atoms needed on the right: $3CO_2$ gives $3\times2=6$ oxygen atoms, and $4H_2O$ gives $4\times1=4$ oxygen atoms, a total of $10$ oxygen atoms, so $5\,O_2$ molecules are needed on the left, already a whole number:

$C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)$

Check: carbon $3=3$; hydrogen $8=4\times2=8$; oxygen $5\times2=10=(3\times2)+(4\times1)=6+4=10$. Balanced.

Part (b): Complete combustion of ethane

Start with the unbalanced skeleton equation: $C_2H_6 + O_2 \rightarrow CO_2 + H_2O$

Balance carbon first (2 carbon atoms in ethane need 2 $CO_2$): $C_2H_6 + O_2 \rightarrow 2CO_2 + H_2O$

Balance hydrogen next (6 hydrogen atoms in ethane need 3 $H_2O$): $C_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O$

Now count oxygen atoms needed on the right: $2CO_2$ gives $2\times2=4$ oxygen atoms, and $3H_2O$ gives $3\times1=3$ oxygen atoms, a total of $7$ oxygen atoms, so $3.5\,O_2$ molecules are needed. Since equations must use whole-number coefficients, double every coefficient in the equation:

$2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)$

Check: carbon $2\times2=4=4$; hydrogen $2\times6=12=6\times2$; oxygen $7\times2=14=(4\times2)+(6\times1)=8+6=14$. Balanced.

Part (c): The general combustion equation for any alkane

Alkanes have general formula $C_nH_{2n+2}$. Following exactly the same method as parts (a) and (b), but using $n$ instead of a specific number of carbon atoms:

Skeleton equation: $C_nH_{2n+2} + O_2 \rightarrow CO_2 + H_2O$

Balance carbon (each of the $n$ carbon atoms needs one $CO_2$): $C_nH_{2n+2} + O_2 \rightarrow nCO_2 + H_2O$

Balance hydrogen (the $(2n+2)$ hydrogen atoms need $(n+1)$ $H_2O$, since each $H_2O$ uses 2 hydrogen atoms): $C_nH_{2n+2} + O_2 \rightarrow nCO_2 + (n+1)H_2O$

Count the oxygen atoms needed on the right: $nCO_2$ gives $2n$ oxygen atoms, and $(n+1)H_2O$ gives $(n+1)$ oxygen atoms, a total of $2n+(n+1)=3n+1$ oxygen atoms, so $\frac{3n+1}{2}\,O_2$ molecules are needed on the left. Because $3n+1$ is not always even (it depends on whether $n$ is odd or even, as seen by comparing part (a), where no doubling was needed, with part (b), where it was), the general equation must be written with whole-number coefficients by doubling every term:

$2C_nH_{2n+2}(g) + (3n+1)O_2(g) \rightarrow 2nCO_2(g) + (2n+2)H_2O(g)$

Check: carbon $2n=2n$; hydrogen $2(2n+2)=4n+4=2(2n+2)$; oxygen $(3n+1)\times2=6n+2=(2n\times2)+((2n+2)\times1)=4n+2n+2=6n+2$. Balanced for any value of $n$.

Consistency check with parts (a) and (b): substituting $n=3$ gives $2C_3H_8+10O_2\rightarrow6CO_2+8H_2O$, which is simply the part (a) equation with every coefficient doubled; substituting $n=2$ gives $2C_2H_6+7O_2\rightarrow4CO_2+6H_2O$, exactly the part (b) equation.

Final answers

• (a) $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)$
• (b) $2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)$
• (c) $2C_nH_{2n+2}(g) + (3n+1)O_2(g) \rightarrow 2nCO_2(g) + (2n+2)H_2O(g)$