Question

Ammonium nitrate, a common fertilizer, was used by terrorists in the tragic explosion in Oklahoma City in \(1995 .\) How many liters of gas at \(307^{\circ} \mathrm{C}\) and 1.00 atm are formed by the explosive decomposition of \(15.0 \mathrm{~kg}\) of ammonium nitrate to nitrogen, oxygen, and water vapor?

Short answer

31,327 liters

Step-by-step solution

- Determine the Decomposition Reaction

Ammonium nitrate (\text{NH}_4\text{NO}_3) decomposes following the balanced chemical equation: \[ 2 \text{NH}_4\text{NO}_3(s) \rightarrow 2 \text{N}_2(g) + O_2(g) + 4 \text{H}_2\text{O}(g) \]

- Calculate the Molar Mass of Ammonium Nitrate

The molar mass of ammonium nitrate (\text{NH}_4\text{NO}_3) is calculated as follows:\[ \text{NH}_4\text{NO}_3 = 14 + (4 \times 1) + 14 + (3 \times 16) = 80 \text{ g/mol} \]

- Convert Mass to Moles

Calculate the number of moles of ammonium nitrate in 15.0 kg:\[ 15,000 \text{ g} \times \frac{1 \text{ mol}}{80 \text{ g}} = 187.5 \text{ mol} \]

- Use Stoichiometry to Determine Moles of Products

From the balanced equation, 2 moles of \text{NH}_4\text{NO}_3 produce 7 moles of gas (2 moles of \text{N}_2, 1 mole of \text{O}_2, and 4 moles of \text{H}_2\text{O}). Therefore, the moles of gas produced by 187.5 moles of \text{NH}_4\text{NO}_3 is:\[ 187.5 \text{ mol} \times \frac{7 \text{ mol gas}}{2 \text{ mol} \text{NH}_4 \text{NO}_3} = 656.25 \text{ mol} \text{ gas} \]

- Apply Ideal Gas Law

Use the ideal gas law equation to determine the volume of gas produced at 307°C (580 K) and 1.00 atm:\[ PV = nRT \]Given: \[ n = 656.25 \text{ mol}, R = 0.0821 \text{ L atm K}^{-1} \text{mol}^{-1}, T = 580 \text{ K}, P = 1.00 \text{ atm} \]\[ V = \frac{nRT}{P} = \frac{656.25 \times 0.0821 \times 580}{1.00} = 31,327 \text{ L} \]

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry. It assists us in understanding the relationship between the pressure, volume, temperature, and number of moles of a gas. The equation is expressed as:
\[ PV = nRT \]
Here's what each term means:

• P stands for pressure in atm (atmospheres).
• V stands for volume in liters (L).
• n stands for the number of moles of the gas.
• R is the ideal gas constant, which is 0.0821 L·atm/K·mol.
• T stands for the temperature in Kelvin (K).

To use this equation, we must ensure all units match up. For instance, temperature must be in Kelvin, not Celsius. Conversion from Celsius to Kelvin is simple: add 273.15.
For example, a temperature of 307°C becomes
307 + 273.15 = 580.15 K.
When you have values for P, n, R, and T, you can find the volume V by rearranging the equation: \[ V = \frac{nRT}{P} \].
We use this rearranged equation to determine the volume of gas formed in the ammonium nitrate decomposition: \[ V = \frac{656.25 * 0.0821 * 580}{1.00} = 31,327 \text{ L} \].

Stoichiometry

Stoichiometry is the part of chemistry that deals with the calculation of reactants and products in chemical reactions. It is key to understand quantities in balanced chemical equations.
Take the decomposition of ammonium nitrate as an example: \[ 2 \text{NH}_4\text{NO}_3(s) \rightarrow 2 \text{N}_2(g) + O_2(g) + 4 \text{H}_2\text{O}(g) \].
This equation tells us that 2 moles of \( \text{NH}_4\text{NO}_3 \) produce 7 moles of gas (2 moles \( \text{N}_2 \), 1 mole \( \text{O}_2 \), and 4 moles \( \text{H}_2\text{O} \)). This mole ratio is key for stoichiometric calculations.
For instance, given 15.0 kg of \( \text{NH}_4\text{NO}_3 \), we convert the mass to moles: \[ 15,000 \text{ g} * \frac{1 \text{ mol}}{80 \text{ g}} = 187.5 \text{ mol} \].
From our balanced equation, we know that 2 moles of \( \text{NH}_4\text{NO}_3 \) produce 7 moles of gas. Thus, \[ 187.5 \text{ mol NH}_4\text{NO}_3 * \frac{7 \text{ mol gas}}{2 \text{ mol NH}_4\text{NO}_3} = 656.25 \text{ mol gas} \].
Using this ratio, we calculate the volume of gas produced.

Mole Calculations

Mole calculations are fundamental in chemistry. The mole is a unit that represents \( 6.022 \times 10^{23} \) particles of a substance, also known as Avogadro's number. This helps chemists count tiny particles like atoms and molecules by weighing them.
To illustrate, let's calculate the moles of ammonium nitrate. First, find its molar mass: \[ \text{NH}_4\text{NO}_3 = 14 + (4 \times 1) + 14 + (3 \times 16) = 80 \text{ g/mol} \].
Next, convert the given mass to moles: \[ \text{mass} / \text{molar mass} = 15,000 \text{ g} / 80 \text{ g/mol} = 187.5 \text{ mol} \].
With 187.5 moles of \( \text{NH}_4\text{NO}_3 \), we use stoichiometry to find the moles of gas produced. The balanced equation shows that 2 moles of \( \text{NH}_4\text{NO}_3 \) produce 7 moles of gas. Thus,
\( 187.5 \text{ mol NH}_4\text{NO}_3 * \frac{7 \text{ mol gas}}{2 \text{ mol NH}_4\text{NO}_3} = 656.25 \text{ mol gas} \).
Understanding mole calculations makes solving chemical problems easier! Learning to convert masses to moles and using mole ratios is key for mastering chemistry.