Question

1 d: practice problems - balance each equation using the process from part c.

challenge problem: give it your best shot!

Explanation:

Step1: Balance carbon atoms

In \(C_2H_6 + O_2
ightarrow CO_2 + H_2O\), for carbon, we have 2 carbon atoms on the left (in \(C_2H_6\)) and 1 on the right (in \(CO_2\)). So we put a coefficient of 2 in front of \(CO_2\). The equation becomes \(C_2H_6+O_2
ightarrow 2CO_2 + H_2O\).

Step2: Balance hydrogen atoms

There are 6 hydrogen atoms on the left (in \(C_2H_6\)) and 2 on the right (in \(H_2O\)). So we put a coefficient of 3 in front of \(H_2O\). The equation is now \(C_2H_6+O_2
ightarrow 2CO_2 + 3H_2O\).

Step3: Balance oxygen atoms

On the right - hand side, we have \(2\times2 + 3\times1=4 + 3 = 7\) oxygen atoms (from \(2CO_2\) and \(3H_2O\)). On the left - hand side, we have \(O_2\). To get 7 oxygen atoms, we can write the coefficient of \(O_2\) as \(\frac{7}{2}\), but we usually prefer whole - number coefficients. So we multiply the entire equation by 2.

Multiplying \(C_2H_6+O_2
ightarrow 2CO_2 + 3H_2O\) by 2 gives \(2C_2H_6+7O_2
ightarrow 4CO_2 + 6H_2O\).

Answer:

\(2C_2H_6+7O_2
ightarrow 4CO_2 + 6H_2O\)