Question

When ammonium carbonate is heated, three gases are produced by its decomposition. $$ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightarrow 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ What total volume of gas is produced, measured at \(453^{\circ} \mathrm{C}\) and 1.04 atm, if \(52.0 \mathrm{g}\) of ammonium carbonate is heated?

Short answer

When 52.0 g of ammonium carbonate is heated at \(453^{\circ} \mathrm{C}\) and 1.04 atm, the total volume of gases produced (ammonia, carbon dioxide, and water vapor) is 144.55 L.

Step-by-step solution

Calculate moles of ammonium carbonate

To find the amount of gases produced, we must begin by finding out how many moles of ammonium carbonate we have. The molar mass of ammonium carbonate \((\mathrm{NH}_{4})_{2} \mathrm{CO}_{3}\) can be calculated using the molar masses of its constituent elements: Molar mass of \(\mathrm{NH}_{4}\) = 14 (N) + 4 x 1 (H) = 18 g/mol Molar mass of \(\mathrm{CO}_{3}\) = 12 (C) + 3 x 16 (O) = 60 g/mol So, the molar mass of \((\mathrm{NH}_{4})_{2} \mathrm{CO}_{3}\) = 2 x 18 + 60 = 96 g/mol. We are given 52.0 g of ammonium carbonate, so to find out how many moles we have, we use the formula: Moles of ammonium carbonate = mass / molar mass = 52.0 g / 96 g/mol = \(0.5417\) moles

Use stoichiometry to find moles of each gas produced

Now, we will use stoichiometry to find the number of moles of each gas produced. From the balanced equation, we have: 1 mole of \((\mathrm{NH}_{4})_{2} \mathrm{CO}_{3}\) produces 2 moles of \(\mathrm{NH}_{3}\), 1 mole of \(\mathrm{CO}_{2}\), and 1 mole of \(\mathrm{H}_{2} \mathrm{O}\) So, moles of \(\mathrm{NH}_{3}\) = 2 x moles of \((\mathrm{NH}_{4})_{2} \mathrm{CO}_{3}\) = 2 x 0.5417 = 1.0834 moles Moles of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) = moles of \((\mathrm{NH}_{4})_{2} \mathrm{CO}_{3}\) = 0.5417 moles each.

Use the ideal gas equation to calculate the volume of each gas

The ideal gas equation is given by: \[PV = nRT\] where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. We are given the pressure (P = 1.04 atm) and the temperature in Celsius (T = 453°C). Convert the temperature to Kelvin: \(T = 453 + 273.15 = 726.15 K\) Choose the correct value for the ideal gas constant, R. Since the pressure is given in atm, we will use R = 0.0821 L·atm/mol·K. Now let's calculate the volume for each gas: \(V_{NH_3}=\frac{n_{NH_3}RT}{P}=\frac{1.0834 \times 0.0821 \times 726.15}{1.04}=72.27 L\) \(V_{CO_2}=\frac{n_{CO_2}RT}{P}=\frac{0.5417 \times 0.0821 \times 726.15}{1.04}=36.14 L\) \(V_{H_2O}=\frac{n_{H_2O}RT}{P}=\frac{0.5417 \times 0.0821 \times 726.15}{1.04}=36.14 L\)

Find the total volume of all gases

Add the volumes of each gas to find the total volume: Total volume = \(V_{NH_3} + V_{CO_2} + V_{H_2O} = 72.27 + 36.14 + 36.14 = 144.55 L\) So the total volume of gases produced when 52.0 g of ammonium carbonate is heated at 453°C and 1.04 atm is 144.55 L.

Key concepts

Stoichiometry

The study of the quantitative relationships, or ratios, between the reactants and products in chemical reactions is known as stoichiometry. It is essential for predicting the outcomes of reactions and for understanding how different chemicals will react with one another. The key to solving a stoichiometry problem is a balanced chemical equation, where the coefficients represent the molar ratios in which substances react and are produced.

For example, in the decomposition of ammonium carbonate, the balanced chemical equation is: \
\( (\mathrm{NH}_4)_2\mathrm{CO}_3(s) \rightarrow 2\mathrm{NH}_3(g) + \mathrm{CO}_2(g) + \mathrm{H}_2O(g) \)
From this equation, we can see that for every mole of ammonium carbonate that decomposes, two moles of ammonia, one mole of carbon dioxide, and one mole of water vapor are produced. These ratios allow us to calculate the number of moles of each gas produced from a given amount of ammonium carbonate.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry and physics that describes the behavior of an ideal gas. It relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas through the equation: \
\[PV = nRT\]
where R is the ideal gas constant. The ideal gas law can be used to calculate the volume occupied by a gas at a given temperature and pressure, provided the number of moles is known.

In our exercise involving the decomposition of ammonium carbonate, the ideal gas law allows us to calculate the volume of each gas produced when we know the number of moles of the gas (from stoichiometry), the temperature in Kelvin, and the pressure of the system. The calculated volumes of ammonia, carbon dioxide, and water vapor can then be summed to find the total volume of gases produced under the specified conditions.

Molar Mass Calculation

Molar mass is defined as the mass of one mole of a substance, and it is measured in grams per mole (g/mol). It is an important concept for converting between the mass of a substance and the number of moles, as well as for stoichiometry calculations. To calculate the molar mass, we sum the masses of all the atoms in the formula of the compound based on the atomic masses found on the periodic table.

In the given exercise, the molar mass of ammonium carbonate \
\((\mathrm{NH}_4)_2\mathrm{CO}_3\) is calculated by adding the mass contributions of the nitrogen (N), hydrogen (H), carbon (C), and oxygen (O) atoms. Knowing the molar mass and the mass of ammonium carbonate provided, we can then calculate the number of moles of ammonium carbonate. This initial step is crucial as it provides the starting point for all subsequent stoichiometry and ideal gas law calculations.