Methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) is a clean-burning fuel. $$\begin{array}{l}{\text { a. Write a balanced chemical equation for }} \\\ {\text { the synthesis of methanol from carbon }} \\ {\text { monoxide and hydrogen gas. }} \\ {\text { b. Write a balanced chemical equation for }} \\\ {\text { the complete combustion of methanol. }}\end{array} $$

Carbon monoxide and hydrogen gas are combined to form methanol. So the unbalanced equation would be \(\mathrm{CO} + \mathrm{H}_{2} \rightarrow \mathrm{CH}_{3}\mathrm{OH}\).

Balance the chemical equation via coefficients. Notice that on the reactant side, there are 2 hydrogen atoms while on the product side, there are 4 hydrogen atoms. Therefore, add a coefficient of 2 for hydrogen gas on the left side to balance it. This leads to \(\mathrm{CO} + 2\mathrm{H}_{2} \rightarrow \mathrm{CH}_{3}\mathrm{OH}\). The equation is now balanced.

Methanol burns in the presence of oxygen to produce carbon dioxide and water. The unbalanced chemical equation would be \(\mathrm{CH}_{3}\mathrm{OH} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\).

First, balance carbons: There's 1 carbon in both sides so it is balanced. Then, balance hydrogens: We have 4 hydrogen atoms on reactant side (from methanol) and 2 on product side (from water). Adding a coefficient of 2 before water on the right side would balance the hydrogen atoms. Now, there are 4 oxygen atoms on the product side (2 from \(\mathrm{CO}_{2}\) and 2 from \(\mathrm{H}_{2}\mathrm{O}\)) and only 2 oxygen atoms on the reactant side. Therefore, balance the oxygen atoms by adding a coefficient of 2 before oxygen on the left side. This results in the balanced chemical equation: \(\mathrm{CH}_{3}\mathrm{OH} + \(\frac{3}{2}\)\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O}\). However, to avoid fractions, multiply the entire equation by 2 to get \(2\mathrm{CH}_{3}\mathrm{OH} + 3\mathrm{O}_{2} \rightarrow 2\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O}\).