Question

Ammonium nitrate has been used as a high explosive because it is unstable and decomposes into several gaseous substances. The rapid expansion of the gaseous substances produces the explosive force. $$ \mathrm{NH}_{4} \mathrm{NO}_{3}(s) \rightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Calculate the mass of each product gas if \(1.25 \mathrm{~g}\) of ammonium nitrate reacts.

Short answer

When 1.25 grams of ammonium nitrate decomposes, it produces approximately 0.4367 grams of nitrogen gas, 0.4992 grams of oxygen gas, and 0.2811 grams of water vapor.

Step-by-step solution

Find the molar mass of ammonium nitrate

In order to convert the mass of ammonium nitrate to moles, we will need its molar mass. Using the periodic table, we can find the molar mass of each element involved in the compound, and then add them up. NH₄NO₃ has the composition of 2 nitrogen (N) atoms, 4 hydrogen (H) atoms, and 3 oxygen (O) atoms. Molar mass of N = 14.01 g/mol Molar mass of H = 1.01 g/mol Molar mass of O = 16.00 g/mol Molar mass of NH₄NO₃ = (2 × 14.01) + (4 × 1.01) + (3 × 16.00) = 28.02 + 4.04 + 48.00 = 80.06 g/mol

Convert given mass of ammonium nitrate to moles

We'll use the molar mass of ammonium nitrate to convert the given mass (1.25 g) to moles. moles of NH₄NO₃ = (1.25 g) / (80.06 g/mol) = 0.0156 mol

Find the moles of each product gas using stoichiometry

The balanced chemical equation provided is: NH₄NO₃(s) → N₂(g) + O₂(g) + H₂O(g) According to the balanced equation, one mole of NH₄NO₃ produces one mole of N₂, one mole of O₂, and one mole of H₂O. Since we have 0.0156 moles of NH₄NO₃, we will also have 0.0156 moles of each product gas.

Find the molar mass of each product gas

We need to find the molar mass of each product gas to convert the moles to mass. Molar mass of N₂ = 2 × 14.01 g/mol = 28.02 g/mol Molar mass of O₂ = 2 × 16.00 g/mol = 32.00 g/mol Molar mass of H₂O = (2 × 1.01) + 16.00 g/mol = 18.02 g/mol

Convert moles of each product gas to mass

Using the molar mass for each product gas, we can find their respective masses. mass of N₂ = moles of N₂ × molar mass of N₂ = (0.0156 mol) × (28.02 g/mol) = 0.4367 g mass of O₂ = moles of O₂ × molar mass of O₂ = (0.0156 mol) × (32.00 g/mol) = 0.4992 g mass of H₂O = moles of H₂O × molar mass of H₂O = (0.0156 mol) × (18.02 g/mol) = 0.2811 g

Conclusion

When 1.25 grams of ammonium nitrate decomposes, it produces approximately 0.4367 grams of nitrogen gas, 0.4992 grams of oxygen gas, and 0.2811 grams of water vapor.

Key concepts

Chemical Reactions

Chemical reactions are processes in which one or more substances transform into new substances, changing their chemical composition. In the decomposition of ammonium nitrate, for example, the solid compound breaks down into nitrogen gas, oxygen gas, and water vapor. During a chemical reaction, bonds are broken and formed, leading to the rearrangement of atoms. This specific reaction is a decomposition type, where a single compound splits into simpler products. Key indicators of a chemical reaction include:

• The production of gas
• Color changes
• Temperature fluctuations
• Formation of precipitates

Understanding chemical reactions involves studying the reactants (starting materials) and products (substances formed) to ascertain the transformation process and its implications.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by adding the atomic masses of all atoms in a molecule. To find the molar mass of a compound like ammonium nitrate, which has the chemical formula NH₄NO₃, you need to account for the number of each type of atom:

• 2 nitrogen (N) atoms
• 4 hydrogen (H) atoms
• 3 oxygen (O) atoms

The molar mass calculator sums the atomic masses: - Nitrogen: 2 atoms × 14.01 g/mol = 28.02 g/mol - Hydrogen: 4 atoms × 1.01 g/mol = 4.04 g/mol - Oxygen: 3 atoms × 16.00 g/mol = 48.00 g/mol
The total molar mass of NH₄NO₃ becomes 28.02 + 4.04 + 48.00 = 80.06 g/mol. This value is crucial for converting between grams and moles, making stoichiometric calculations possible.

Balancing Equations

Balancing equations ensures that the same number of each type of atom appears on both sides of the chemical equation. It is an application of the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction.
In the decomposition of ammonium nitrate: \[\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \rightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\] Each element is balanced:

• 2 nitrogen atoms in both reactants and products
• 4 hydrogen atoms transformed into 2 water molecules
• 3 oxygen atoms forming 1 molecule each of oxygen and water

Balancing involves adjusting coefficients in front of chemical formulas to equalize atom counts without changing the actual formulas themselves. This balanced equation reflects that one mole of ammonium nitrate decomposes to produce specific amounts of nitrogen, oxygen, and water vapor.

Gas Calculations

Gas calculations involve determining the quantity of gases involved in a reaction in terms of moles, mass, or volume. For reactions at standard temperature and pressure (STP), one mole of any gas occupies 22.4 liters.
Given the decomposition of ammonium nitrate, you calculate the mass of gases produced from a known quantity of the reactant. With 1.25 g of ammonium nitrate, you first determine the moles using its molar mass (80.06 g/mol):

• 0.0156 mol of NH₄NO₃

This translates directly into moles of the product gases due to the one-to-one mole ratio in the balanced equation. Finally, using the molar masses:

• Mass of N₂: 28.02 g/mol × 0.0156 mol = 0.4367 g
• Mass of O₂: 32.00 g/mol × 0.0156 mol = 0.4992 g
• Mass of H₂O: 18.02 g/mol × 0.0156 mol = 0.2811 g

Gas calculations permit the accurate prediction of reactants and products in gaseous form, essential in fields from chemistry to engineering.