Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) is also called "laughing gas." \(\mathrm{It}\) can be prepared by the thermal decomposition of ammonium nitrate \(\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)\). The other product is \(\mathrm{H}_{2} \mathrm{O} .\) (a) Write a balanced equation for this reaction. (b) How many grams of \(\mathrm{N}_{2} \mathrm{O}\) are formed if 0.46 mole of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) is used in the reaction?

To understand any chemical reaction, it's essential to start with a balanced chemical equation. This means that the number of atoms for each element is the same on both the reactant and product sides of the equation. In the thermal decomposition of ammonium nitrate \(\mathrm{NH}_4\mathrm{NO}_3\) to produce nitrous oxide \(\mathrm{N}_2\mathrm{O}\) and water \(\mathrm{H}_2\mathrm{O}\), we start with the unbalanced equation: \[\mathrm{NH}_4\mathrm{NO}_3 \rightarrow \mathrm{N}_2\mathrm{O} + \mathrm{H}_2\mathrm{O}\].

Upon inspection, we find that this equation is already balanced as is, with the count of nitrogen, hydrogen, and oxygen atoms being equal on both sides. Each molecule of ammonium nitrate decomposes to give one molecule of nitrous oxide and two molecules of water without needing additional coefficients. It's a straightforward equation, but not all reactions are this simple to balance. In more complex cases, balancing involves iteratively adjusting coefficients in front of compounds to equalize the atom count across the reaction.

Stoichiometry is akin to a recipe for chemistry, dictating the precise amounts of reactants needed to produce desired products. In the context of our example, stoichiometry tells us that one mole of ammonium nitrate \(\mathrm{NH}_4\mathrm{NO}_3\) will produce one mole of nitrous oxide \(\mathrm{N}_2\mathrm{O}\) when it decomposes. This 1:1 ratio is vital because it allows us to predict the amount of product formed from a given amount of reactant.

For instance, if you start with 0.46 moles of \(\mathrm{NH}_4\mathrm{NO}_3\), stoichiometry indicates that you'll produce an equivalent 0.46 moles of \(\mathrm{N}_2\mathrm{O}\). This direct relationship simplifies the process of conversion between moles of different substances within a chemical reaction. Here, understanding the stoichiometric ratio is a key step in solving problems related to the amounts of reactants and products in a chemical reaction.

Molar mass is the bridge that connects the microscopic world of atoms and molecules to the macroscopic world we can measure. It represents the mass of one mole of a substance and is expressed in grams per mole (g/mol). For nitrous oxide \(\mathrm{N}_2\mathrm{O}\), the molar mass is the sum of the atomic masses of nitrogen and oxygen, derived from the periodic table, resulting in approximately 44.01 g/mol.

To calculate the mass of the nitrous oxide produced in our reaction, we use the formula: \[\text{mass} = \text{moles} \times \text{molar mass}\]. With 0.46 moles of \(\mathrm{N}_2\mathrm{O}\) produced, multiplying by its molar mass gives us about 20.2 grams of nitrous oxide. Being able to convert moles to grams using molar mass is a fundamental skill in chemistry that enables us to measure out the substances we need for a reaction or to characterize the products we've generated.