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, put a coefficient of 2 in front of \(CO_2\).
\(CH_3CH_3(g)+O_2(g)\to2CO_2(g)+H_2O(g)\)

Step2: Balance hydrogen atoms

There are 6 hydrogen atoms in \(CH_3CH_3\). So, put a coefficient of 3 in front of \(H_2O\).
\(CH_3CH_3(g)+O_2(g)\to2CO_2(g)+3H_2O(g)\)

Step3: Balance oxygen atoms

On the right - hand side, there are \(2\times2 + 3\times1=7\) oxygen atoms. So, put a coefficient of \(\frac{7}{2}\) in front of \(O_2\).
\(CH_3CH_3(g)+\frac{7}{2}O_2(g)\to2CO_2(g)+3H_2O(g)\)
Since we need whole - number coefficients, multiply each coefficient by 2.
\(2CH_3CH_3(g)+7O_2(g)\to4CO_2(g)+6H_2O(g)\)

Answer:

\(2CH_3CH_3(g)+7O_2(g)\to4CO_2(g)+6H_2O(g)\)