Question

Calculate how many moles of \(\mathrm{NH}_{3}\) form when each quantity of reactant completely reacts. $$3 \mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow 4 \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g)$$ a. 2.6 \(\mathrm{mol} \mathrm{N}_{2} \mathrm{H}_{4}\) b. 3.55 \(\mathrm{mol} \mathrm{N}_{2} \mathrm{H}_{4}\) c. 65.3 \(\mathrm{g} \mathrm{N}_{2} \mathrm{H}_{4}\) d. 4.88 \(\mathrm{kg} \mathrm{N}_{2} \mathrm{H}_{4}\)

Short answer

a. 3.47 moles NH_3, b. 4.73 moles NH_3, c. 2.72 moles NH_3, d. 203.33 moles NH_3

Step-by-step solution

Understanding the Chemical Reaction

The chemical equation given is: 3 N_2H_4(l) -> 4 NH_3(g) + N_2(g). This tells us that 3 moles of hydrazine (N_2H_4) react to form 4 moles of ammonia (NH_3). We'll use this stoichiometric ratio to find out the amount of NH_3 produced from a given amount of N_2H_4.

Calculating Moles for (a)

For 2.6 moles of N_2H_4, use the stoichiometric ratio from the balanced equation to find moles of NH_3. Since 3 moles of N_2H_4 produce 4 moles of NH_3, we calculate the moles of NH_3 by multiplying 2.6 moles of N_2H_4 by the ratio 4/3:(2.6 moles N_2H_4) * (4 moles NH_3 / 3 moles N_2H_4) = 3.47 moles NH_3.

Calculating Moles for (b)

For 3.55 moles of N_2H_4, again use the stoichiometric ratio to find moles of NH_3:(3.55 moles N_2H_4) * (4 moles NH_3 / 3 moles N_2H_4) = 4.73 moles NH_3.

Calculating Moles for (c)

First, we need to convert mass of N_2H_4 to moles using its molar mass. The molar mass of N_2H_4 is 32 g/mol. Thus, for 65.3 grams, we have: (65.3 g N_2H_4) / (32 g/mol) = 2.04 moles N_2H_4. Next, we find moles of NH_3 using the stoichiometric ratio:(2.04 moles N_2H_4) * (4 moles NH_3 / 3 moles N_2H_4) = 2.72 moles NH_3.

Calculating Moles for (d)

First, convert the mass of N_2H_4 from kilograms to grams: (4.88 kg) * (1000 g/kg) = 4880 g. Next, convert mass to moles: (4880 g) / (32 g/mol) = 152.5 moles N_2H_4. Then use the stoichiometric ratio to find moles of NH_3: (152.5 moles N_2H_4) * (4 moles NH_3 / 3 moles N_2H_4) = 203.33 moles NH_3.

Key concepts

Chemical Reaction Calculations

Chemical reaction calculations are crucial for estimating the amounts of substances involved in chemical reactions. Understanding these calculations requires recognizing the relationship between reactants and products as presented in the balanced chemical equation.

In our exercise, the balanced equation \(3 \mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow 4 \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g)\) reveals the stoichiometric ratio: 3 moles of hydrazine \(\mathrm{N}_{2}\mathrm{H}_{4}\) yield 4 moles of ammonia \(\mathrm{NH}_{3}\). By converting given quantities of a reactant to moles, if not already provided, we can then use this ratio to calculate the amount of product formed.

To enhance students' grasp of this concept, it's helpful to work through diverse examples, showing how the conversion factors derive from the coefficients of the balanced equation. An example emphasizing this is the provided Step 2, where we determine moles of ammonia from a known quantity of hydrazine using the stoichiometric relation.

Mole Concept

The mole concept is a fundamental principle in chemistry that quantifies substances. A mole represents Avogadro's number (approximately \(6.022 \times 10^{23}\) entities), whether atoms, molecules, or particles.

In our context, when the problem asks for the number of moles of \(\mathrm{NH}_{3}\) that form, we are being asked to figure out how many Avogadro's worth of ammonia molecules will result from the given reactant quantities. For instance, in Step 2, we used the mole concept to understand that from 2.6 moles of \(\mathrm{N}_{2}\mathrm{H}_{4}\), we can derive the quantity of ammonia produced through mathematical calculation.

Clarifying the mole concept with real-world analogies, such as comparing moles to 'dozens', helps students relate the abstract nature of 'mole' to more familiar counting units. Emphasizing the mole as a bridge between the microscopic scale of atoms and the macroscopic world we interact with can further solidify understanding.

Molar Mass

Molar mass, the weight of one mole of a substance, is typically expressed in grams per mole (g/mol). It's pivotal in converting between the mass of a substance and the number of moles.

In the exercise, to answer part c, we first had to determine how many moles of hydrazine \(\mathrm{N}_{2}\mathrm{H}_{4}\) were present in 65.3 grams. Knowing the molar mass of hydrazine, 32 g/mol, made this possible. By dividing the mass by the molar mass, we arrived at the mole quantity required for further stoichiometric calculations.

For students, understanding molar mass as the link between the mass that can be measured on a scale and moles that govern chemical reactions is crucial. Illustrated through visual aids, like diagrams comparing the mass of different moles of substances, and exercises converting grams to moles, can strengthen the connection between theory and practice.