For the reaction shown, calculate how many moles of \(\mathrm{NH}_{3}\) form when each amount of reactant completely reacts. $$ 3 \mathrm{~N}_{2} \mathrm{H}_{4}(l) \longrightarrow 4 \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g) $$ (a) \(5.3 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\) (b) \(2.28 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\) (c) \(5.8 \times 10^{-2} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\) (d) \(9.76 \times 10^{7} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\)

Understanding chemical reaction calculations is a fundamental part of chemistry that allows scientists and students to predict the outcomes of chemical reactions. In essence, these calculations involve using the balanced chemical equation as a recipe for determining how much of each reactant is needed to produce a desired amount of product.

For example, let's consider the reaction provided in the exercise: \[3 \text{N}_2\text{H}_4(l) \rightarrow 4 \text{NH}_3(g) + \text{N}_2(g)\]. Here, the equation tells us that three moles of hydrazine (\(\text{N}_2\text{H}_4\)) react to form four moles of ammonia (\(\text{NH}_3\)) and nitrogen gas (\(\text{N}_2\)). In a real-world scenario, chemists use such equations to calculate the amounts of reactants required or products formed during the reaction.
By mastering chemical reaction calculations, students learn to comprehend and apply concepts such as conservation of mass and stoichiometry, enabling them to solve practical problems efficiently.

A pivotal aspect of stoichiometry is the mole ratio, which is derived from the coefficients of a balanced chemical equation. It gives us the proportional relationship between reactants and products. For the balanced chemical reaction in our exercise, the mole ratio between hydrazine and ammonia is \(3:4\).

This ratio means that for every three moles of \(\text{N}_2\text{H}_4\) that react, four moles of \(\text{NH}_3\) are produced. Understanding the mole ratio is essential because it serves as a conversion factor, allowing one to calculate how many moles of a product are made from a given amount of reactant or vice versa. Emphasizing this concept aids students in visualizing the quantitative aspect of chemical reactions and reinforces the idea that molecules react in fixed proportions, not individual amounts.

Knowing how to determine the moles of reactant and product in a chemical equation is crucial for predicting the quantities involved in the reaction. A mole is a unit of measurement for amount of substance, similar to how a dozen refers to a quantity of twelve.

In the exercise, we look at different amounts of \(\text{N}_2\text{H}_4\) ranging from a few moles to a very large number. Each case illustrates how the initial amount of reactant dictates the amount of product formed. For example, 5.3 moles of \(\text{N}_2\text{H}_4\) will produce a different amount of \(\text{NH}_3\) compared to 2.28 moles of the same reactant. By explaining and providing examples on how to calculate these amounts, we ensure students can transfer these skills to various chemical reactions and real-world chemical processes.

The ultimate goal of stoichiometric calculations is to quantify the relationships in a chemical reaction. These calculations are firmly based on the principle of conservation of matter; in a closed system, matter is neither created nor destroyed. Stoichiometry uses the mole ratio as a key tool for quantitative analysis of reactants and products.

In each step of the provided exercise, the stoichiometric calculations involve multiplying the moles of \(\text{N}_2\text{H}_4\) by the mole ratio obtained from the balanced chemical equation to find the moles of \(\text{NH}_3\). This conversion is simple in theory but requires careful attention to detail to ensure the correct ratios are used and the units cancel accordingly. Teaching students to perform such calculations with precision prepares them for more complex problems they may encounter in both academic and industrial chemistry settings.