Question

The energy used to power one of the Apollo lunar missions was supplied by the following overall reaction: \(2 \mathrm{N}_{2} \mathrm{H}_{4}+\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2}+3 \mathrm{N}_{2} \mathrm{O}_{4} \longrightarrow\) \(6 \mathrm{N}_{2}+2 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}\) . For the phase of the mission when the lunar module ascended from the surface of the moon, a total of \(1200 . \mathrm{kg} \mathrm{N}_{2} \mathrm{H}_{4}\) was available to react with \(1000 . \mathrm{kg}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2}\) and \(4500 . \mathrm{kg} \mathrm{N}_{2} \mathrm{O}_{4}\) a. For this portion of the flight, which of the allocated components was used up first? b. How much water, in kilograms, was put into the lunar atmosphere through this reaction?

Short answer

The limiting reactant is \(\mathrm{N}_{2} \mathrm{O}_{4}\). The amount of water produced is 2347.83 kg.

Step-by-step solution

Write and Balance the Chemical Equation

The given chemical reaction is:\[2 \mathrm{N}_{2} \mathrm{H}_{4} + \left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2} + 3 \mathrm{N}_{2} \mathrm{O}_{4} \longrightarrow 6 \mathrm{N}_{2} + 2 \mathrm{CO}_{2} + 8 \mathrm{H}_{2} \mathrm{O}\]This equation is already balanced.

Calculate Molar Masses

Calculate the molar masses of each reactant:\[\mathrm{N}_{2} \mathrm{H}_{4}: 2(14) + 4(1) = 32 \mathrm{g/mol}\]\[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2}: 2\left(12 + 3(1) + 14\right) + 2\left(1\right) = 60 \mathrm{g/mol}\]\[\mathrm{N}_{2} \mathrm{O}_{4}: 2(14) + 4(16) = 92 \mathrm{g/mol}\]

Convert Masses to Moles

Convert the given masses into moles:\[\mathrm{N}_{2} \mathrm{H}_{4}: \frac{1200\,\mathrm{kg}}{32\,\mathrm{g/mol}} = \frac{1200 \times 10^3\,\mathrm{g}}{32\,\mathrm{g/mol}} = 37500\,\mathrm{mol}\]\[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2}: \frac{1000\,\mathrm{kg}}{60\,\mathrm{g/mol}} = \frac{1000 \times 10^3\,\mathrm{g}}{60\,\mathrm{g/mol}} = 16666.67\,\mathrm{mol}\]\[\mathrm{N}_{2} \mathrm{O}_{4}: \frac{4500\,\mathrm{kg}}{92\,\mathrm{g/mol}} = \frac{4500 \times 10^3\,\mathrm{g}}{92\,\mathrm{g/mol}} = 48913.04\,\mathrm{mol}\]

Determine Limiting Reactant

Using the balanced equation proportion, calculate the amount of each reactant needed:For \(\mathrm{N}_{2} \mathrm{H}_{4} \to \frac{37500}{2} = 18750\,\text{reaction units}\)For \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2} \to \frac{16666.67}{1} = 16666.67\,\text{reaction units}\)For \(\mathrm{N}_{2} \mathrm{O}_{4} \to \frac{48913.04}{3} = 16304.35\,\text{reaction units}\)The limiting reactant is \(\mathrm{N}_{2} \mathrm{O}_{4}\) because it has the smallest number of reaction units.

Calculate Water Produced

Using the limiting reactant (\(\mathrm{N}_{2} \mathrm{O}_{4}\)), calculate the amount of water produced. The reaction shows that 3 moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\) produce 8 moles of \(\mathrm{H}_{2} \mathrm{O}\):\[\text{Moles of } \mathrm{H}_{2} \mathrm{O} = \frac{48913.04}{3} \times 8 = 130434.78\,\text{moles}\]Convert moles of water to kilograms: \(\mathrm{H}_{2} \mathrm{O}\) molar mass is \(18\,\mathrm{g/mol}\).\[\text{Mass of } \mathrm{H}_{2} \mathrm{O} = 130434.78 \times 18\,\mathrm{g} = 2347826.04\,\mathrm{g} = 2347.83\,\mathrm{kg}\]

Key concepts

chemical reaction

A chemical reaction is a process where reactant molecules transform into product molecules. In chemical equations, the substances you start with are called reactants, while the substances formed are called products. The reaction in this problem is:

\[2 \mathrm{N}_{2} \mathrm{H}_{4} + \left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2} + 3 \mathrm{N}_{2} \mathrm{O}_{4} \longrightarrow 6 \mathrm{N}_{2} + 2 \mathrm{CO}_{2} + 8 \mathrm{H}_{2} \mathrm{O}\]
This notation tells us about the substances participating in the reaction and how they convert from reactants into products.

mole conversion

Mole conversion helps us switch between the mass of a substance and its amount in moles. To convert from grams to moles, you use the molar mass of the substance (found from the periodic table). For example, the molar mass of \(\mathrm{N}_{2} \mathrm{H}_{4} \) (hydrazine) is 32 g/mol.

To convert 1200 kg of \(\mathrm{N}_{2} \mathrm{H}_{4} \) to moles, we can use:
\[\text{Moles of } \mathrm{N}_{2} \mathrm{H}_{4} = \frac{1200 \text{ kg} \times 10^3 \text{ g/kg}}{32 \text{ g/mol}} = 37500 \text{ mol}\]
This conversion is crucial for comparing the amounts of different reactants in the reaction.

stoichiometry

Stoichiometry involves using a balanced chemical equation to calculate the quantities of reactants and products. It tells us the stoichiometric ratios in which substances react and are produced. From our balanced equation, we know:

\[2 \mathrm{N}_{2} \mathrm{H}_{4} + \left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2} + 3 \mathrm{N}_{2} \mathrm{O}_{4} \longrightarrow 6 \mathrm{N}_{2} + 2 \mathrm{CO}_{2} + 8 \mathrm{H}_{2} \mathrm{O}\]
This shows that 2 moles of \(\mathrm{N}_{2} \mathrm{H}_{4} \) react with 1 mole of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2} \) and 3 moles of \(\mathrm{N}_{2} \mathrm{O}_{4} \) to produce products like water.

balanced equation

A balanced equation ensures that the number of atoms for each element is the same on both the reactant and product sides, following the law of conservation of mass. In our reaction:

\[2 \mathrm{N}_{2} \mathrm{H}_{4} + \left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2} + 3 \mathrm{N}_{2} \mathrm{O}_{4} \longrightarrow 6 \mathrm{N}_{2} + 2 \mathrm{CO}_{2} + 8 \mathrm{H}_{2} \mathrm{O}\]
The atoms of nitrogen (N), hydrogen (H), carbon (C), and oxygen (O) are balanced. Such equations are foundational for understanding how amounts of each reactant relate to each other and to the products formed. This balance is critical for correctly identifying the limiting reactant in our problem.