Problem 49 - 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 \(20^{\circ} \mathrm{C} .\) What mass of \(\mathrm{N}_{2} \mathrm{H}_{4}\) is needed to consume all of the dissolved \(\mathbf{O}_{2}\) in 3.00 \(\times 10^{4} \mathrm{L}\) of water (enough to fill a small swimming pool)?

Step-by-Step Solution

Answer

1322.06 g of hydrazine is needed.

Step 1: Write the Chemical Equation

The reaction involves hydrazine (_2H_4) reacting with oxygen (O_2) to produce nitrogen gas (N_2) and water. The unbalanced equation is: \( \mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{O}_{2} \rightarrow \mathrm{N}_{2} + \mathrm{H}_{2} \mathrm{O} \).

Step 2: Balance the Equation

To balance the equation, ensure that the number of each type of atom on the reactant side equals the number on the product side. The balanced equation is: \( \mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{O}_{2} \rightarrow \mathrm{N}_{2} + 2 \mathrm{H}_{2} \mathrm{O} \). This balances nitrogen and the hydrogen atoms on both sides.

Step 3: Calculate Mass of Dissolved Oxygen

Use the information given that 0.0044 g of \(\mathrm{O}_{2}\) dissolves in 100 mL of water to find the total mass of oxygen in 30,000 L of water:Convert volume from liters to milliliters:\[ 3.00 \times 10^4 \text{ L} = 3.00 \times 10^7 \text{ mL} \]Use proportionality:\[ \text{Total mass of } \mathrm{O}_{2} = \frac{3.00 \times 10^7}{100} \times 0.0044 = 1320 \text{ g} \].

Step 4: Determine Moles of O_2

Convert the mass of oxygen to moles using its molar mass (32 g/mol):\[ \text{Moles of } \mathrm{O}_{2} = \frac{1320 \text{ g}}{32 \text{ g/mol}} = 41.25 \text{ mol} \].

Step 5: Calculate Amount of Hydrazine Needed

From the balanced equation, 1 mole of \(\mathrm{N}_{2} \mathrm{H}_{4}\) reacts with 1 mole of \(\mathrm{O}_{2}\). Therefore, 41.25 moles of \(\mathrm{N}_{2} \mathrm{H}_{4}\) are needed.Determine the mass of hydrazine using its molar mass (32.05 g/mol):\[ \text{Mass of } \mathrm{N}_{2} \mathrm{H}_{4} = 41.25 \text{ mol} \times 32.05 \text{ g/mol} = 1322.0625 \text{ g} \].

Key Concepts

Chemical Equations

Chemical equations are simply representations of chemical reactions. They show how different substances, known as reactants, interact to form new substances, called products. In a chemical equation, you start with the reactants on the left side, followed by an arrow, and end with the products on the right side.

Balancing a chemical equation is crucial. This means ensuring there is an equal number of each type of atom on both sides of the equation. This process is based on the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a closed system. For example, in the hydrazine and oxygen reaction, we begin with the unbalanced equation: \[\mathrm{N}_{2}\mathrm{H}_{4} + \mathrm{O}_{2} \rightarrow \mathrm{N}_{2} + \mathrm{H}_{2}\mathrm{O}\]

Upon balancing, each side of the equation has equal numbers of the same atoms, giving us:\[\mathrm{N}_{2}\mathrm{H}_{4} + \mathrm{O}_{2} \rightarrow \mathrm{N}_{2} + 2 \mathrm{H}_{2} \mathrm{O}\]This ensures the conservation of both nitrogen and hydrogen atoms.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantities of reactants and products in a chemical reaction. It is a tool used to calculate the amounts of substances consumed and produced during reactions.

In our hydrazine reaction, stoichiometry helps determine the amount of hydrazine required to react with a given quantity of oxygen. The balanced equation indicates that 1 mole of \(\mathrm{N}_{2}\mathrm{H}_{4}\) reacts with 1 mole of \(\mathrm{O}_{2}\), providing a 1:1 molar ratio. Thus, if you have a certain amount of oxygen, the same amount of hydrazine will be necessary.

• Moles of \(\mathrm{O}_{2}\): 41.25 mol
• Moles of \(\mathrm{N}_{2}\mathrm{H}_{4}\) required: 41.25 mol

Such calculations are essential for determining the proper proportions in any chemical process or reaction.

Oxygen Solubility

Oxygen solubility refers to the amount of oxygen that can dissolve in water. This can vary with temperature, pressure, and the presence of other substances. In this context, we see that at 20°C, oxygen has a solubility of 0.0044 g in 100 mL of water.

When tackling problems involving large volumes of water, such as the swimming pool example given, the concept of solubility helps us determine how much oxygen is actually available to react. For 30,000 liters of water, by converting liters to milliliters, we can estimate the total mass of dissolved oxygen:

• 30000 L = 30000000 mL
• Total mass of \(\mathrm{O}_{2}\): \(\frac{30000000}{100} \times 0.0044 = 1320 \text{ g}\)

Understanding oxygen solubility is crucial for reactions that occur in aquatic environments.

Molecular Mass Calculations

Molecular mass calculations are used to determine the mass of one mole of a chemical compound. This is done by summing the atomic masses of all atoms in the compound formula.

In the exercise, hydrazine \(\mathrm{N}_{2}\mathrm{H}_{4}\) has a molar mass calculated from the atomic masses of nitrogen (N) and hydrogen (H):

• Atomic mass of N: 14.01 g/mol
• Atomic mass of H: 1.008 g/mol

To get the molar mass of hydrazine:
\[32.05 \text{ g/mol} = (2 \times 14.01) + (4 \times 1.008)\]

These calculations provide insight into the macroscopic quantities needed for chemical reactions. Determining the mass of reactants or products is a fundamental skill in chemistry, enabling precise and efficient experimental planning.