Question

Dinitrogen oxide, commonly called nitrous oxide, is used as a propellant gas for whipped-cream dispensers. It is prepared by heating ammonium nitrate to \(250^{\circ} \mathrm{C}\). Water vapor is also formed. (a) Write a balanced equation for the decomposition of ammonium nitrate into nitrous oxide and steam. (b) What volume of dinitrogen oxide gas is formed at \(250^{\circ} \mathrm{C}\) and 1,0 atm when \(5.00 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) is heated? Assume \(100 \%\) yield.

Short answer

Answer: 2.65 L

Step-by-step solution

Write the balanced chemical equation

The decomposition reaction of ammonium nitrate can be written as follows: $$\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \rightarrow \mathrm{N}_{2}\mathrm{O}(g) + 2\mathrm{H}_{2}\mathrm{O}(g)$$

Determine moles of ammonium nitrate and the stoichiometry ratio

We are given that \(5.00\,\mathrm{g}\) of ammonium nitrate is heated. First, we need to find the molar mass of ammonium nitrate: $$\mathrm{NH}_{4} \mathrm{NO}_{3} = 1(14.01) + 4(1.01) + 1(14.01) + 3(16.00) = 80.05\,\mathrm{g/mol}$$ Then, we determine the moles of ammonium nitrate: $$\frac{5.00\,\mathrm{g}}{80.05\,\mathrm{g/mol}} = 0.0624\,\mathrm{mol}$$ From the balanced chemical equation, we know that 1 mole of ammonium nitrate (\(\mathrm{NH}_{4}\mathrm{NO}_{3}\)) decomposes to produce 1 mole of dinitrogen oxide (\(\mathrm{N}_{2}\mathrm{O}\)). Therefore, the moles of dinitrogen oxide produced are the same as the moles of ammonium nitrate decomposed: $$0.0624\,\mathrm{mol}\,\mathrm{N}_{2}\mathrm{O}$$

Apply the ideal gas law

We use the ideal gas law formula to find the volume of dinitrogen oxide gas formed at \(250^{\circ}\mathrm{C}\) and \(1.0\,\mathrm{atm}\): $$PV = nRT$$ Rearrange the formula to solve for volume (V): $$V = \frac{nRT}{P}$$ We have: \(n = 0.0624\,\mathrm{mol}\) \(R = 0.0821\,\mathrm{L\cdot atm/mol\cdot K}\) \(T = 250^{\circ}\mathrm{C} + 273.15 = 523.15\,\mathrm{K}\) \(P = 1.0\,\mathrm{atm}\) Plugging in the values, we get the volume: $$V = \frac{(0.0624\,\mathrm{mol})(0.0821\,\mathrm{L\, atm/mol\cdot K})(523.15\,\mathrm{K})}{1.0\,\mathrm{atm}}$$

Calculate the volume of dinitrogen oxide gas

Calculate the value of the expression above: $$V = 2.65\,\mathrm{L}$$ So, when \(5.00\,\mathrm{g}\) of ammonium nitrate is heated at \(250^{\circ} \mathrm{C}\) and \(1.0\,\mathrm{atm}\) pressure, \(2.65\,\mathrm{L}\) of dinitrogen oxide gas is formed.

Key concepts

Balanced Chemical Equation

Understanding how to balance a chemical equation is essential for solving many problems in chemistry. To balance an equation, you must ensure the same number of each type of atom appears on both sides of the reaction arrow, essentially conserving mass as per the Law of Conservation of Mass.

For the decomposition of ammonium nitrate, the correct balanced equation is: \[\mathrm{NH}_{4}\mathrm{NO}_{3}(s) \rightarrow \mathrm{N}_{2}\mathrm{O}(g) + 2\mathrm{H}_{2}\mathrm{O}(g)\]
This translates to one mole of solid ammonium nitrate (\(\mathrm{NH}_{4}\mathrm{NO}_{3}\)) decomposing to yield one mole of gaseous dinitrogen oxide (\(\mathrm{N}_{2}\mathrm{O}\)) and two moles of gaseous water vapor (\(2\mathrm{H}_{2}\mathrm{O}\)). The balancing process takes into account that in this decomposition reaction, nitrogen, hydrogen, and oxygen atoms are rearranged, and the total mass remains constant before and after the reaction.

Stoichiometry

Stoichiometry revolves around the quantitative relationships between the reactants and products in a chemical reaction. Based on the balanced chemical equation, stoichiometry allows one to predict the amount of reactants needed or products formed.

For instance, in our reaction: \[\mathrm{NH}_{4}\mathrm{NO}_{3}(s) \rightarrow \mathrm{N}_{2}\mathrm{O}(g) + 2\mathrm{H}_{2}\mathrm{O}(g)\],
we calculate the moles of ammonium nitrate using its molar mass, which is the sum of the atomic masses of each atom in the molecule, yielding \(80.05\,\mathrm{g/mol}\). Understanding the stoichiometry between ammonium nitrate and dinitrogen oxide tells us that decomposing \(5.00\,\mathrm{g}\) of ammonium nitrate will yield an equivalent molar amount of dinitrogen oxide. This molar ratio is crucial for predicting the extent of a reaction and is a fundamental concept in chemistry helping to connect the microscopic world of molecules to the macroscopic world we can measure.

Ideal Gas Law

The Ideal Gas Law is an essential tool in chemistry, defined by the equation \(PV = nRT\), where \(P\) represents pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. This relationship describes the behavior of ideal gases, assuming gas particles do not attract or repel each other and occupy no volume.

In our problem, we utilize the Ideal Gas Law to determine the volume of dinitrogen oxide produced at a given temperature and pressure. By rearranging the formula to solve for volume (\(V\)) and substituting the known values, including converting the temperature to Kelvin, we can find the volume of dinitrogen oxide gas formed from the decomposed ammonium nitrate. This law is incredibly useful for calculating properties of gases under various conditions and helps in understanding the concept of gas density, molar volume, and conversions between moles and the physical volume of gases.