Consider the reaction between \(\mathrm{N}_{2} \mathrm{H}_{4}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) : $$ 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ A reaction vessel initially contains \(27.5 \mathrm{~g} \mathrm{~N}_{2} \mathrm{H}_{4}\) and \(74.9 \mathrm{~g}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}\). Calculate the masses of \(\mathrm{N}_{2} \mathrm{H}_{4}, \mathrm{~N}_{2} \mathrm{O}_{4}, \mathrm{~N}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\) that will be in the reaction vessel after the reactants have reacted as much as possible. Assume \(100 \%\) yield. Hint: The limiting reactant is completely consumed, but the reactant in excess is not. Use the amount of limiting reactant to determine the amount of products that form and the amount of the reactant in excess that remains after complete reaction.

In chemical reactions, the limiting reactant, or limiting reagent, is the substance which is entirely consumed first and thus determines the amount of product that is formed. Understanding the concept of the limiting reactant is crucial because it dictates the maximum yield of the reaction.

To identify the limiting reactant, one must first know the balanced chemical equation for the reaction. From there, the comparison of reactant mole ratios allows us to determine the reactant that will be consumed first. In our exercise, when looking at the reaction of \( \mathrm{N}_{2} \mathrm{H}_{4}\) and \( \mathrm{N}_{2} \mathrm{O}_{4}\), it's evident that the amounts of each reactant given are not present in the stoichiometric ratio as per the balanced chemical equation. By comparing the quantity in moles of each reactant to their stoichiometric coefficients, we identify \( \mathrm{N}_{2} \mathrm{H}_{4}\) as the limiting reactant.

An important take-home message is that once the limiting reactant is used up, the reaction cannot proceed further even if there is an excess of the other reactants. The remaining reactants are termed excess reactants and remain unconsumed in the reaction mixture.

Molar mass calculation is a fundamental skill required to convert between mass and moles of a substance, which is a key step in stoichiometry problems. Molar mass is defined as the mass of one mole of a particular substance and is typically expressed in grams per mole (g/mol).

To calculate the molar mass of a compound, you need to sum the molar masses of all the atoms that make up the compound, multiplying the atomic masses from the periodic table by the number of each type of atom present in the formula. For example, in our exercise, the molar mass of \( \mathrm{N}_{2} \mathrm{H}_{4}\) is determined by adding twice the molar mass of nitrogen (\(2 \times 14 \text{ g/mol}\)) to four times the molar mass of hydrogen (\(4 \times 1 \text{ g/mol}\)).

This process is crucial for converting grams to moles, which allows us to use the balanced chemical equation to relate quantities of reactants and products. A key point to remember is that accurate determination of molar masses is essential for calculating the correct amounts of reactants and products involved in a chemical reaction.

Chemical reaction equations provide a symbolic representation of a chemical reaction. They detail the reactants—it's what you start with, and the products—it's what you end up with, as well as the quantities of each substance involved, expressed through their stoichiometric coefficients.

To make use of chemical reaction equations effectively, it's crucial to ensure that the equation is balanced. This means that the number of atoms of each element involved remains consistent before and after the reaction. This reflects the law of conservation of mass. In our exercise, the balanced equation \(2 \mathrm{~N}_{2} \mathrm{H}_{4}(g) + \mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 3 \mathrm{~N}_{2}(g) + 4 \mathrm{H}_{2} \mathrm{O}(g)\) indicates the exact ratio in which reactants combine and the ratio in which products form.

Once students understand how to read and write chemical equations, they can delve deeper into predicting the outcomes of reactions, which includes calculating theoretical yields and identifying the extent to which reactants are consumed. Chemical reaction equations are the roadmap of stoichiometry; they allow us to calculate not only the amount of products formed in a reaction but also, as shown in our example, the remaining amount of excess reactants.