Sulfur dioxide and water are formed when hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), reacts with oxygen gas. Write the balanced equation for the reaction.

First, write down the unbalanced chemical equation, which represents the reaction between hydrogen sulfide (H\(_2\)S) and oxygen gas (O\(_2\)) to form sulfur dioxide (SO\(_2\)) and water (H\(_2\)O): H\(_2\)S + O\(_2\) \(\rightarrow\) SO\(_2\) + H\(_2\)O

Now, we need to balance the chemical equation by adjusting the coefficients (the numbers in front of the chemical formulas) so that the number of atoms for each element is equal on both the reactant and product sides of the equation. 1. Balance the sulfur atoms: Since there is 1 sulfur atom in both H\(_2\)S and SO\(_2\), we place a coefficient of 1 in front of SO\(_2\): H\(_2\)S + O\(_2\) \(\rightarrow\) 1SO\(_2\) + H\(_2\)O 2. Balance the hydrogen atoms: There are 2 hydrogen atoms in H\(_2\)S and 2 hydrogen atoms in H\(_2\)O. Since there is already a coefficient of 1 in front of H\(_2\)O, we don't need to change anything: H\(_2\)S + O\(_2\) \(\rightarrow\) SO\(_2\) + 1H\(_2\)O 3. Balance the oxygen atoms: There are 2 oxygen atoms in O\(_2\) and a total of 3 oxygen atoms in the products (2 in SO\(_2\) and 1 in H\(_2\)O). To obtain an even number of oxygen atoms on both sides, we can double the number of H\(_2\)S and H\(_2\)O molecules by placing a coefficient of 2 in front of H\(_2\)S and H\(_2\)O: 2H\(_2\)S + O\(_2\) \(\rightarrow\) 2SO\(_2\) + 2H\(_2\)O By doing this, we also balance the sulfur atoms and hydrogen atoms: - There are 4 hydrogen atoms on both sides (2 in each H\(_2\)S molecule and 2 in each H\(_2\)O molecule). - There are 2 sulfur atoms on both sides (1 in each H\(_2\)S molecule and 1 in each SO\(_2\) molecule). Now the chemical equation is balanced: 2H\(_2\)S + O\(_2\) \(\rightarrow\) 2SO\(_2\) + 2H\(_2\)O