Question

Rocket Fuel The exothermic reaction between liquid hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{2}\right)\) and liquid hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) is used to fuel rockets. The products of this reaction are nitrogen gas and water. a. Write the balanced chemical equation. b. How much hydrazine, in grams, is needed to produce 10.0 mol of nitrogen gas?

Short answer

a. The balanced chemical equation for the reaction between liquid hydrazine and liquid hydrogen peroxide is: \(N_2H_4\) + 2\(H_2O_2\) → \(N_2\) + 4\(H_2O\) b. 320.4 g of hydrazine is needed to produce 10.0 mol of nitrogen gas.

Step-by-step solution

Balance the Chemical Equation

To write the balanced chemical equation for this reaction, we first need to write the chemical formula of both reactants and products: \(N_2H_4\) (Hydrazine) + \(H_2O_2\) (Hydrogen peroxide) → \(N_2\) (Nitrogen gas) + \(H_2O\) (Water) Next, we will balance this chemical equation by adding stoichiometric coefficients to ensure that the same number of each type of atom appears on both sides of the equation: \(N_2H_4\) + 2\(H_2O_2\) → \(N_2\) + 4\(H_2O\)

Calculate Moles of Hydrazine Needed

To calculate the moles of hydrazine needed to produce 10.0 mol of nitrogen gas, we can use the balanced chemical equation to determine the molar relationship between hydrazine and nitrogen gas: 1 mol \(N_2H_4\) → 1 mol \(N_2\) Since the ratio between hydrazine and nitrogen gas is 1:1, we require 10.0 mol of hydrazine to produce 10.0 mol of nitrogen gas.

Calculate Mass of Hydrazine Needed

To find the mass of hydrazine needed, we will use the formula: Mass = moles × molar mass First, find the molar mass of hydrazine (\(N_2H_4\)): \(N_2H_4\) → (2 × 14.01 g/mol) + (4 × 1.01 g/mol) ≈ 32.04 g/mol Now, multiply the moles of hydrazine required (10.0 mol) by the molar mass of hydrazine (32.04 g/mol): Mass of hydrazine required = (10.0 mol) × (32.04 g/mol) = 320.4 g Therefore, 320.4 g of hydrazine is needed to produce 10.0 mol of nitrogen gas.

Key concepts

Balanced Chemical Equations

Understanding balanced chemical equations is fundamental in the study of chemistry, especially when analyzing reactions, such as the one used in rocket fuel. But, what does it mean to balance a chemical equation? Essentially, it’s applying the Law of Conservation of Mass, which states that matter cannot be created or destroyed in an isolated system. In chemical reactions, this law translates to having an equal number of each type of atom on both the reactant side and the product side of the equation.

For the reaction between hydrazine ((N_2H_4)) and hydrogen peroxide ((H_2O_2)) yielding nitrogen gas ((N_2)) and water ((H_2O)), balancing assures us we do not lose any atoms during the conversion. We add coefficients to adjust the number of molecules to ensure that for every atom of nitrogen and hydrogen on the left, there's an identical number on the right. A balanced equation also enables us to understand the stoichiometry of the reaction — the ratio in which the reactants combine to form products.

Stoichiometry

Stoichiometry is essentially the 'recipe' for chemistry; it involves the quantitative relationship between reactants and products in a chemical reaction. It relies heavily on balanced chemical equations because these ratios are necessary for any calculation involving the substances in a reaction.

In our rocket fuel example, the stoichiometry tells us that for every molecule of (N_2H_4) used, one molecule of (N_2) is produced. This is a 1:1 molar ratio. Knowing this, we can then determine how many moles of hydrazine are required to produce a certain amount of nitrogen gas. If the problem states we need 10.0 mol of (N_2), stoichiometry allows us to calculate that we'll also need 10.0 mol of (N_2H_4). It is this straightforward understanding that allows chemists to predict the outcomes of reactions and prepare the right amounts of each reactant.

Molar Mass Calculations

Molar mass is a critical component when converting between moles and grams - a process often required in stoichiometry. Every substance has a unique molar mass, equivalent to its gram molecular weight per mole. It is calculated by summing the atomic masses of all the atoms in a molecule, as per the periodic table.

Taking the hydrazine molecule ((N_2H_4)), for instance, we calculate its molar mass by adding the masses of nitrogen and hydrogen atoms. With a molar mass of approximately 32.04 g/mol for hydrazine, we can then easily convert moles to grams. For 10.0 mol of hydrazine, multiplying by the molar mass (32.04 g/mol) gives us 320.4 g. This process of using molar mass calculations is essential for accurately measuring the amounts of substances needed in a laboratory or industrial setting, especially when precise amounts of reactants are crucial, as in the preparation of rocket fuel.