Question

balance the chemical equation below using the smallest possible whole number stoichiometric coefficients. ch₃ch₃(g) + o₂(g) → co₂(g) + h₂o(g)

Explanation:

Step1: Balance carbon atoms

There are 2 carbon atoms in $CH_3CH_3$. So, we put a 2 in front of $CO_2$:
$CH_3CH_3(g)+O_2(g)
ightarrow2CO_2(g)+H_2O(g)$

Step2: Balance hydrogen atoms

There are 6 hydrogen atoms in $CH_3CH_3$. So, we put a 3 in front of $H_2O$:
$CH_3CH_3(g)+O_2(g)
ightarrow2CO_2(g)+3H_2O(g)$

Step3: Balance oxygen atoms

On the right - hand side, there are $2\times2 + 3\times1=7$ oxygen atoms. So, we put $\frac{7}{2}$ in front of $O_2$:
$CH_3CH_3(g)+\frac{7}{2}O_2(g)
ightarrow2CO_2(g)+3H_2O(g)$
To get whole - number coefficients, we multiply each coefficient by 2:
$2CH_3CH_3(g)+7O_2(g)
ightarrow4CO_2(g)+6H_2O(g)$

Answer:

$2CH_3CH_3(g)+7O_2(g)
ightarrow4CO_2(g)+6H_2O(g)$