Problem 65 - Chemistry and Chemical Reactivity

Question

A major use of hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) is in steam boilers in power plants. (a) The reaction of hydrazine with \(\mathrm{O}_{2}\) dissolved in water gives \(\mathrm{N}_{2}\) and water. Write a balanced equation for this reaction. (b) \(\mathrm{O}_{2}\) dissolves in water to the extent of \(0.0044 \mathrm{g}\) in \(100 .\) mL of water at \(20^{\circ}\) C. What mass of \(\mathrm{N}_{2} \mathrm{H}_{4}\) is needed to consume all of the dissolved \(\mathrm{O}_{2}\) in \(3.00 \times 10^{4} \mathrm{L}\) of water (enough to fill a small swimming pool)?

Step-by-Step Solution

Answer

1321.31 g of \(\mathrm{N}_{2}\mathrm{H}_{4}\) is required.

Step 1: Understand the Reaction

The chemical reaction involves hydrazine (\(\mathrm{N}_{2} \mathrm{H}_{4}\)) reacting with dissolved oxygen (\(\mathrm{O}_{2}\)) to produce nitrogen gas (\(\mathrm{N}_{2}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)).

Step 2: Write the Unbalanced Reaction Equation

The unbalanced chemical equation describing the reaction is: \(\mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{O}_{2} \rightarrow \mathrm{N}_{2} + \mathrm{H}_{2}\mathrm{O}\).

Step 3: Balance the Chemical Equation

Adjust the coefficients to balance the number of each type of atom on both sides of the reaction. The balanced chemical equation is: \[\mathrm{N}_{2}\mathrm{H}_{4} + \mathrm{O}_{2} \rightarrow \mathrm{N}_{2} + 2\mathrm{H}_{2}\mathrm{O}.\]

Step 4: Calculate Oxygen Mass for Given Volume

Use the solubility information to calculate the mass of \(\mathrm{O}_{2}\) in \(3.00 \times 10^4\) L of water. \[ \text{Mass of } \mathrm{O}_{2} = 0.0044\, \text{g/100 mL} = 0.0044\, \text{g/0.1 L}. \] Convert it to the total volume: \[ \text{Total mass of } \mathrm{O}_{2} = 0.0044\, \text{g/0.1 L} \times 3.00 \times 10^4 \text{ L} = 1320\, \text{g}. \]

Step 5: Use Reaction Stoichiometry

The balanced equation shows \(1\) mole of \(\mathrm{N}_{2}\mathrm{H}_{4}\) reacts with \(1\) mole of \(\mathrm{O}_{2}\). Calculate moles of \(\mathrm{O}_{2}\): \[ \text{Moles of } \mathrm{O}_{2} = \frac{1320\, \text{g}}{32.00\, \text{g/mol}} = 41.25\, \text{moles}. \] You need the same moles of \(\mathrm{N}_{2}\mathrm{H}_{4}\).

Step 6: Calculate Mass of Hydrazine Needed

Given \(41.25\) moles of \(\mathrm{N}_{2}\mathrm{H}_{4}\) are required, calculate the mass: \[ \text{Molar mass of } \mathrm{N}_{2} \mathrm{H}_{4} = 32.05\, \text{g/mol}. \] Calculate: \[ \text{Mass of } \mathrm{N}_{2}\mathrm{H}_{4} = 41.25\, \text{mol} \times 32.05\, \text{g/mol} = 1321.31\, \text{g}. \]

Step 7: Conclusion: Mass of Hydrazine

The total mass of \(\mathrm{N}_{2}\mathrm{H}_{4}\) required to react with all dissolved \(\mathrm{O}_{2}\) is approximately \(1321.31\, \text{g}.\)

Key Concepts

Reaction Stoichiometry

In chemical reactions, stoichiometry is all about the relationships between the quantities of reactants and products in a chemical reaction. It is essentially the recipe of chemistry that you need to follow in order to transform reactants into products. When dealing with stoichiometry, it's crucial to understand the mole concept, as chemical equations are often expressed in moles.
To determine how much of a substance is required or produced, you need to use the stoichiometric coefficients in the balanced chemical equation. These coefficients tell you the ratio in which the reactants combine and how much product they will form.
For instance, in the provided exercise, one mole of hydrazine reacts with one mole of oxygen, forming products. This tells us in the stoichiometric calculations that for every mole of oxygen consumed, one mole of hydrazine is needed. This fundamental concept helps in planning the amounts needed for industrial processes or in a laboratory setting.

Balancing Chemical Equations

Balancing chemical equations is a foundational skill in chemistry that ensures the law of conservation of mass is upheld. This means that in any chemical reaction, the mass of the reactants must be equal to the mass of the products.
To balance an equation, you need equal numbers of each type of atom on both sides of the chemical equation. Let's take the unbalanced equation from the exercise: \[\mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{O}_{2} \rightarrow \mathrm{N}_{2} + \mathrm{H}_{2}\mathrm{O}\]This equation is unbalanced because the number of hydrogens and oxygens are not the same on both sides. For example, there are four hydrogen atoms in one molecule of hydrazine, but only two in one molecule of water. By adjusting the coefficients, you get:\[\mathrm{N}_{2}\mathrm{H}_{4} + \mathrm{O}_{2} \rightarrow \mathrm{N}_{2} + 2\mathrm{H}_{2}\mathrm{O}\]Now the equation is balanced with four hydrogens and two oxygens on both sides. Balancing equations is like solving a puzzle, making sure every atom counts.

Solubility in Water

Solubility refers to the ability of a substance to dissolve in a solvent, such as water. This property is crucial in chemical reactions, especially those in aqueous solutions. The solubility of a gas like oxygen in water can be affected by temperature among other factors.
In our scenario, the solubility of oxygen (\(\mathrm{O}_{2}\)) in water is given as \(0.0044 \mathrm{g}\) in \(100\, \mathrm{mL}\), or roughly \(0.0044 \mathrm{g/0.1 \mathrm{L}}\). This information allows us to calculate how much oxygen is in a larger volume of water, such as in a swimming pool.
The calculation involves converting that value to the necessary larger scale, taking into account the total volume of water. This step is essential in ensuring that we have an accurate figure of how much reactant is required, particularly when dealing with large bodies of water.

Molar Mass Calculations

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (\(g/mol\)). It is a fundamental concept in stoichiometry because it helps convert between grams and moles, the two key units in chemical calculations.
To calculate the mass of reactants needed or products produced, you start by knowing the molar mass of each substance. For example, the molar mass of hydrazine (\(\mathrm{N}_{2} \mathrm{H}_{4}\)) is calculated by adding the atomic masses of its constituent atoms: 2 nitrogen atoms and 4 hydrogen atoms, giving \(32.05\, \mathrm{g/mol}\).
In the exercise, we're required to calculate the mass of hydrazine needed for a specific reaction. Since we know the number of moles from the stoichiometry and the molar mass, we can easily compute the needed mass: \[\mathrm{Mass} = \mathrm{Moles} \times \mathrm{Molar Mass}\]Understanding molar mass is crucial in bridging the gap between measurable mass and the more theoretical mole concept used in balanced chemical equations.