Problem 84 - Chemistry and Chemical Reactivity

Question

Ammonia gas is synthesized by combining hydrogen and nitrogen: $$ 3 \mathrm{H}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g}) $$ (a) If you want to produce 562 g of \(\mathrm{NH}_{3}\), what volume of \(\mathrm{H}_{2}\) gas, at \(56^{\circ} \mathrm{C}\) and \(745 \mathrm{mm} \mathrm{Hg}\), is required? (b) To produce 562 g of \(\mathrm{NH}_{3},\) what volume of air (the source of \(\mathrm{N}_{2}\) ) is required if the air is introduced at \(29^{\circ} \mathrm{C}\) and \(745 \mathrm{mm}\) Hg? (Assume the air sample has \(\left.78.1 \text { mole } \% N_{2} .\right)\)

Step-by-Step Solution

Answer

You need 1374.6 L of H2 and 534.4 L of air.

Step 1: Calculate moles of NH3

The molar mass of ammonia, \(\mathrm{NH}_3\), is \(14.01 + 3(1.008) = 17.034\) g/mol. Calculate the moles of \(\mathrm{NH}_3\) by using the mass given.\[\text{Moles of } \mathrm{NH}_3 = \frac{562 \text{ g}}{17.034 \text{ g/mol}} \approx 33.00 \text{ mol}\]

Step 2: Use stoichiometry to find moles of H2

The balanced chemical equation shows that 2 moles of \(\mathrm{NH}_3\) are produced for every 3 moles of \(\mathrm{H}_2\). Using the moles of \(\mathrm{NH}_3\) calculated, we find moles of \(\mathrm{H}_2\).\[\text{Moles of } \mathrm{H}_2 = \frac{3 \text{ moles H}_2}{2 \text{ moles NH}_3} \times 33.00 \text{ moles NH}_3 = 49.5 \text{ moles H}_2\]

Step 3: Determine volume of H2 gas

Apply the ideal gas law \(PV = nRT\) to find the volume of \(\mathrm{H}_2\) gas required. Given the pressure \(P = 745 \text{ mmHg} = 0.980 \text{ atm}\), \(T = 56^\circ C = 329 \text{ K}\) and \(n = 49.5\) moles.\[V = \frac{nRT}{P} = \frac{49.5 \times 0.0821 \times 329}{0.980}\]After solving, \(V \approx 1374.6\text{ L}\)

Step 4: Calculate moles of N2 needed

Since 2 moles of \(\mathrm{NH}_3\) are produced from 1 mole of \(\mathrm{N}_2\), calculate moles of \(\mathrm{N}_2\) needed:\[\text{Moles of } \mathrm{N}_2 = \frac{1 \text{ moles N}_2}{2 \text{ moles NH}_3} \times 33.00 \text{ moles NH}_3 = 16.5 \text{ moles N}_2\]

Step 5: Calculate volume of air for N2

Since air is composed of 78.1% \(\mathrm{N}_2\) by mole, divide the moles of \(\mathrm{N}_2\) by 0.781 to find total moles of air:\[\text{Total moles of air} = \frac{16.5}{0.781} \approx 21.13 \text{ moles air}\]Apply the ideal gas law assuming conditions \(P = 745 \text{ mmHg} = 0.980 \text{ atm}\), \(T = 29^\circ C = 302 \text{ K}\):\[V = \frac{21.13 \times 0.0821 \times 302}{0.980} \approx 534.4 \text{ L}\]

Key Concepts

Ideal Gas Law

The ideal gas law is a mathematical formula used to describe the behavior of gases. It's a crucial tool in chemistry for understanding how gases will react under different conditions. The formula for the ideal gas law is given by \(PV = nRT\). Let's break it down to understand it better:

• \(P\) is the pressure of the gas, usually measured in atmospheres (atm) or millimeters of mercury (mmHg).
• \(V\) is the volume the gas occupies, which is typically measured in liters (L).
• \(n\) represents the number of moles of gas, which ties into how much gas you have.
• \(R\) is the ideal gas constant, often approximated as 0.0821 L·atm/mol·K.
• \(T\) is the temperature in Kelvin; remember to convert Celsius to Kelvin by adding 273.15 to the Celsius temperature.

The ideal gas law is essential when dealing with gases because it allows us to predict how a gas will respond when one of the conditions changes (like pressure or temperature). It is particularly helpful for combining gases in chemical reactions, such as the synthesis of ammonia, where we need to calculate the volume of hydrogen gas required to produce a certain amount of ammonia.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In our exercise, we're looking at the synthesis of ammonia (\(\mathrm{NH}_3\)) from hydrogen (\(\mathrm{H}_2\)) and nitrogen (\(\mathrm{N}_2\)). The balanced chemical equation for this reaction is:\[3 \mathrm{H}_{2} (g) + \mathrm{N}_{2} (g) \rightarrow 2 \mathrm{NH}_{3} (g)\]This equation shows the stoichiometry of the reaction, meaning it illustrates the exact proportions in which the reactants combine to form the products.

• The numbers in front of the chemical formulas, like the 3 in front of \(\mathrm{H}_2\), represent the moles of each component involved in the reaction.
• It tells us that 3 moles of hydrogen gas react with 1 mole of nitrogen gas to produce 2 moles of ammonia gas.

Understanding the stoichiometry in a chemical reaction is fundamental for calculating how much of each reactant is needed or how much product will be formed. It's critical when performing laboratory experiments or industrial synthesis like producing ammonia for fertilizers.

Mole Calculations

Mole calculations are an integral part of stoichiometry in chemistry, helping us quantify the amount of substances involved in reactions. The mole is a standard unit in chemistry that measures the number of particles, typically atoms or molecules, in a sample.The steps to perform mole calculations usually involve:

• Determining the molar mass of compounds by adding up the atomic masses of the elements found on the periodic table.
• Using the formula \(\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}\) to convert the mass of a substance into moles.
• Applying stoichiometry for converting moles of reactants to moles of products, as indicated in the balanced chemical equation.

In our ammonia synthesis example, we calculated the moles of \(\mathrm{NH}_3\) produced from a given mass, and then used stoichiometry to find out how many moles of \(\mathrm{H}_2\) and \(\mathrm{N}_2\) were needed. This detailed process is crucial for ensuring that chemical reactions are carried out with the correct proportions, leading to efficient use of resources and minimizing waste.