Question

In Problem 6.40 , hydrazine reacted with oxygen according to the following (unbalanced) equation: $$\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ (a) If \(75.0 \mathrm{~kg}\) of hydrazine are reacted with \(75.0 \mathrm{~kg}\) of oxygen, which is the limiting reagent? (b) How many kilograms of \(\mathrm{NO}_{2}\) are produced from the reaction of \(75.0 \mathrm{~kg}\) of the limiting reagent? (c) If \(59.3 \mathrm{~kg}\) of \(\mathrm{NO}_{2}\) are obtained from the reaction in part (a), what is the percent yield?

Short answer

(a) \(\mathrm{O}_2\) is the limiting reagent. (b) 107.8 kg \(\mathrm{NO}_2\) is produced. (c) Percent yield is 55.0\%.

Step-by-step solution

Balance the Chemical Equation

Start by balancing the chemical equation for the reaction of hydrazine and oxygen. The unbalanced equation is:\[\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\]To balance it, use the coefficients: 1 for \(\mathrm{N}_2\mathrm{H}_4\), 2 for \(\mathrm{O}_2\), 2 for \(\mathrm{NO}_2\), and 2 for \(\mathrm{H}_2\mathrm{O}\). The balanced equation is:\[\mathrm{N}_{2} \mathrm{H}_{4} + 2 \mathrm{O}_{2} \rightarrow 2 \mathrm{NO}_{2} + 2 \mathrm{H}_{2} \mathrm{O}\]

Calculate Moles of Reactants

Calculate the number of moles of \(\mathrm{N}_{2} \mathrm{H}_{4}\) and \(\mathrm{O}_{2}\). Use the respective molar masses: \(\mathrm{N}_2\mathrm{H}_4 = 32.05 \text{ g/mol}\) and \(\mathrm{O}_2 = 32.00 \text{ g/mol}\).For \(75 \text{ kg} = 75,000 \text{ g}\):- Moles of \(\mathrm{N}_{2} \mathrm{H}_{4}\): \[ \text{moles of } \mathrm{N}_{2} \mathrm{H}_{4} = \frac{75,000}{32.05} \approx 2339.6 \text{ moles}\]- Moles of \(\mathrm{O}_{2}\): \[ \text{moles of } \mathrm{O}_{2} = \frac{75,000}{32.00} = 2343.75 \text{ moles}\]

Determine Limiting Reagent

Based on the balanced equation, 1 mole of \(\mathrm{N}_{2} \mathrm{H}_{4}\) reacts with 2 moles of \(\mathrm{O}_{2}\). Therefore, the stoichiometric ratio is 1:2.Calculate how much \(\mathrm{O}_{2}\) is required for \(\mathrm{N}_{2} \mathrm{H}_{4}\):\[2339.6 \times 2 = 4679.2 \text{ moles of } \mathrm{O}_{2}\]Since you only have 2343.75 moles of \(\mathrm{O}_{2}\), \(\mathrm{O}_{2}\) is the limiting reagent.

Calculate NO2 Produced

Determine moles of \(\mathrm{NO}_{2}\) produced using limiting reagent \(\mathrm{O}_{2}\).Balanced equation shows 2 moles of \(\mathrm{NO}_{2}\) are produced from 2 moles of \(\mathrm{O}_{2}\).Moles of \(\mathrm{NO}_{2}\):\[\text{moles of } \mathrm{NO}_{2} = 2343.75 \text{ moles (same as } \mathrm{O}_{2}\text{)}\]Convert moles of \(\mathrm{NO}_{2}\) to kilograms:\[\text{molar mass of } \mathrm{NO}_{2} = 46.01 \text{ g/mol}\]\[\text{mass of } \mathrm{NO}_{2} = 2343.75 \times 46.01 / 1000 \approx 107.8 \text{ kg}\]

Calculate Percent Yield

Use the formula for percent yield:\[\text{Percent Yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\%\]Here, the actual yield is \(59.3 \text{ kg}\) and the theoretical yield is \(107.8 \text{ kg}\):\[\text{Percent Yield} = \left( \frac{59.3}{107.8} \right) \times 100 \approx 55.0\%\]

Key concepts

Percent Yield

When performing a chemical reaction, the actual amount of product may vary from the calculated theoretical yield. This variance is expressed as percent yield. To calculate percent yield, you compare the actual yield (what was actually obtained from the experiment) with the theoretical yield (the maximum possible amount predicted by stoichiometry). The formula for percent yield is:

• Percent Yield = \( \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \)

In the exercise above, the actual yield of \( \mathrm{NO}_{2} \) is 59.3 kg, while the theoretical yield, calculated using stoichiometry, is 107.8 kg. Thus, the percent yield is approximately 55.0%.
This indicates that only 55% of the expected product was obtained, possibly due to side reactions, incomplete reactions, or other experimental errors.

Balancing Chemical Equations

Balancing chemical equations is crucial in chemistry as it ensures that the law of conservation of mass is upheld. This principle states that matter cannot be created or destroyed in a chemical reaction. Therefore, the number of atoms of each element must be equal on both the reactant and product sides of the equation.
To balance an equation, adjust the coefficients in front of each compound, without changing the chemical formulas themselves. For the hydrazine and oxygen reaction:

• Unbalanced: \(\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)
• Balanced: \(\mathrm{N}_{2} \mathrm{H}_{4} + 2 \mathrm{O}_{2} \rightarrow 2 \mathrm{NO}_{2} + 2 \mathrm{H}_{2} \mathrm{O}\)

This balanced equation now represents that 1 mole of hydrazine reacts with 2 moles of oxygen, forming 2 moles each of nitrogen dioxide and water vapor.

Stoichiometry

Stoichiometry is the process of using balanced chemical equations to calculate the quantities of reactants and products in a chemical reaction. It allows chemists to predict how much of each substance will be consumed or produced.
In the exercise, we first determine the moles of each reactant. Using their molar masses, the number of moles of hydrazine is about 2339.6, and for oxygen, it is approximately 2343.75.
Next, we identify the limiting reagent, which is the reactant that determines the maximum yield of product. Based on the balanced equation for hydrazine and oxygen, 2 moles of oxygen are needed for each mole of hydrazine. This indicates that oxygen is the limiting reagent because more moles of it are needed than are available.
Finally, the stoichiometric calculations allow us to predict the moles, and thus the mass, of \( \mathrm{NO}_{2} \) produced. A total of 107.8 kg of \( \mathrm{NO}_{2} \) is predicted to form, assuming complete reaction of both reactants without losses or side reactions.