91Ó°ÊÓ

Steering rockets on the space shuttle are powered by the reaction of hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\), and dinitrogen tetraoxide, \(\mathrm{N}_{2} \mathrm{O}_{4}\). The reactions for making hydrazine and the reaction with dinitrogen tetraoxide are as follows: $$ \begin{array}{c} 2 \mathrm{NH}_{3}(a q)+\mathrm{NaOCl}(a q) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(a q)+\mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \\ 2 \mathrm{~N}_{2} \mathrm{H}_{4}(l)+\mathrm{N}_{2} \mathrm{O}_{4}(l) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) \end{array} $$ Starting with \(50.0 \mathrm{~mL}\) of \(6.00 \mathrm{M}\) ammonia and excess other reactants, calculate: (a) the mass of nitrogen gas produced (b) the volume of nitrogen gas produced at STP (c) the mass of water produced

Step by step solution

01

Calculate Moles of Ammonia

Given 50.0 mL of 6.00 M ammonia, first convert the volume from mL to L: \(50.0 \text{ mL} = 0.0500 \text{ L}\). Use the molarity to find the moles: \(\text{moles of } \mathrm{NH}_3 = 6.00 \text{ M} \times 0.0500 \text{ L} = 0.300 \text{ moles NH}_3\).

02

Calculate Moles of Hydrazine Produced

Based on the equation for making hydrazine: \(2\mathrm{NH}_{3}(aq) + \mathrm{NaOCl}(aq) \rightarrow \mathrm{N}_2\mathrm{H}_4(aq) + \mathrm{NaCl}(aq) + \mathrm{H}_2\mathrm{O}(l)\), 2 moles of \(\mathrm{NH}_3\) produce 1 mole of \(\mathrm{N}_2\mathrm{H}_4\). So, the moles of \(\mathrm{N}_2\mathrm{H}_4\) produced = \(\frac{0.300}{2} = 0.150 \text{ moles N}_2\mathrm{H}_4\).

03

Calculate Moles of Nitrogen Gas Produced

Using the reaction \(2\mathrm{N}_2\mathrm{H}_4(l) + \mathrm{N}_2\mathrm{O}_4(l) \rightarrow 3\mathrm{N}_2(g) + 4\mathrm{H}_2\mathrm{O}(g)\), the molar ratio of \(\mathrm{N}_2\mathrm{H}_4\) to \(\mathrm{N}_2\) is 2:3. Thus, \(0.150 \text{ moles } \mathrm{N}_2\mathrm{H}_4\) produce \(0.150 \times \frac{3}{2} = 0.225 \text{ moles } \mathrm{N}_2\).

04

Calculate Mass of Nitrogen Gas Produced

The molar mass of \(\mathrm{N}_2\) is \(28.02 \text{ g/mol}\). Therefore, the mass = \(0.225 \text{ moles} \times 28.02 \text{ g/mol} = 6.305 \text{ g}\).

05

Calculate Volume of Nitrogen Gas at STP

At STP, 1 mole of gas occupies 22.4 L. Therefore, the volume of \(0.225 \text{ moles } \mathrm{N}_2\) is \(0.225 \times 22.4 = 5.04 \text{ L}\).

06

Calculate Moles of Water Produced

From the reaction \(2\mathrm{N}_2\mathrm{H}_4 + \mathrm{N}_2\mathrm{O}_4 \rightarrow 3\mathrm{N}_2 + 4\mathrm{H}_2\mathrm{O}\), the molar ratio of \(\mathrm{N}_2\mathrm{H}_4\) to \(\mathrm{H}_2\mathrm{O}\) is 2:4. So, \(0.150 \text{ moles } \mathrm{N}_2\mathrm{H}_4\) produce \(0.150 \times \frac{4}{2} = 0.300 \text{ moles of } \mathrm{H}_2\mathrm{O}\).

07

Calculate Mass of Water Produced

The molar mass of \(\mathrm{H}_2\mathrm{O}\) is \(18.02 \text{ g/mol}\). Therefore, the mass of water is \(0.300 \text{ moles} \times 18.02 \text{ g/mol} = 5.406 \text{ g}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions

Chemical reactions are processes in which substances transform into new substances. They involve the breaking and forming of chemical bonds, which is highlighted by chemical equations. In our exercise, two main reactions occur:

• Hydrazine (\(\mathrm{N}_2\mathrm{H}_4\)) is formed when ammonia (\(\mathrm{NH}_3\)) reacts with sodium hypochlorite (\(\mathrm{NaOCl}\)).
• This hydrazine then reacts with dinitrogen tetraoxide (\(\mathrm{N}_2\mathrm{O}_4\)) to produce nitrogen gas (\(\mathrm{N}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)).

Understanding these equations can help you foresee the products formed and determine the quantities required or produced. Coefficients in the balanced equation indicate the ratio of moles of reactants and products, key to calculating how much of each substance is involved.

Molarity

Molarity is a measurement of the concentration of a solution, expressed as moles of solute per liter of solution. It is represented with the unit M (mol/L). When dealing with chemical reactions in solutions, knowing the molarity helps calculate the moles of reactants or products.
To find the number of moles using molarity, multiply the molarity by the volume of the solution in liters. For example, if you have 6.00 M ammonia and a volume of 50.0 mL, you convert the volume to liters (0.0500 L) and multiply: \(\text{moles of \(\mathrm{NH}_3\)} = 6.00 \, \text{M} \times 0.0500 \, \text{L} = 0.300 \, \text{moles}\).

• Helps in determining the starting amount of reactants.
• Essential for stoichiometry calculations.

Gas Laws

Gas laws describe how gases behave under various conditions of pressure, volume, and temperature. One critical concept is STP (Standard Temperature and Pressure), where conditions are set at 0°C (273.15 K) and 1 atm pressure. Under STP, one mole of a gas occupies 22.4 liters.
In the exercise, the volume of nitrogen gas produced can be calculated using its amount in moles and its behavior under STP:

• Using the relation: \(\text{Volume} = \text{moles} \times 22.4 \, \text{L/mol}\)
• Example for nitrogen gas: \(0.225 \, \text{moles} \times 22.4 \, \text{L/mol} = 5.04 \, \text{L}\).

Understanding these calculations is important when predicting the behavior of gases in chemical reactions and real-life conditions, allowing chemists to design and modify systems such as rocket engines efficiently.