Question

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$$ a. What is the maximum mass of ammonia that can be produced from a mixture of \(1.00 \times 10^{3} \mathrm{g} \mathrm{N}_{2}\) and \(5.00 \times 10^{2} \mathrm{g} \mathrm{H}_{2} ?\) b. What mass of which starting material would remain unreacted?

Short answer

a. The maximum mass of ammonia that can be produced is \(1.22 \times 10^{3} g\). b. The starting material that would remain unreacted is hydrogen, with a mass of \(2.84 \times 10^{2} g\).

Step-by-step solution

Write down the given information

In this exercise, we have a balanced chemical reaction: \[\mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\] The given mass of Nitrogen (N2): \(1.00 \times 10^{3} g\) The given mass of Hydrogen (H2): \(5.00 \times 10^{2} g\)

Calculate the moles of starting material

To begin, we need to convert the given mass of each starting material (N2 and H2) into moles. Molar mass of N2 = 2 * 14.007 = 28.014 g/mol Molar mass of H2 = 2 * 1.008 = 2.016 g/mol Now, we can calculate the moles of nitrogen and hydrogen using the mass and molar mass: Moles of N2 = \(\frac{1.00 \times 10^{3} g}{28.014 \frac {g}{mol}}\) = 35.7 mol Moles of H2 = \(\frac{5.00 \times 10^{2} g}{2.016 \frac {g}{mol}}\) = 248 mol

Calculate moles of ammonia produced and find the limiting reagent

Given the balanced chemical equation, we can determine the stoichiometric ratio between the moles of N2, H2, and NH3: N2:H2:NH3 = 1:3:2 Next, let's calculate the moles of ammonia produced from each starting material based on the ratio: Moles of NH3 produced from N2 = 35.7 mol * (2 mol NH3 / 1 mol N2) = 71.4 mol NH3 Moles of NH3 produced from H2 = 248 mol * (2 mol NH3 / 3 mol H2) = 165.3 mol NH3 The limiting reagent is the one that produces the least amount of product (NH3). In this case, the limiting reagent is N2.

Calculate the maximum mass of ammonia that can be produced

As N2 is the limiting reagent, we use the moles of ammonia produced from N2 to determine the maximum mass of ammonia that can be produced: Molar mass of NH3 = (14.007 + 3 * 1.008) = 17.031 g/mol Max mass of NH3 = 71.4 mol * 17.031 g/mol = 1.22 x 10^3 g

Calculate the mass of unreacted starting material

From the stoichiometric ratio, we can determine the amount of H2 that would have reacted with the 35.7 mol of N2: Moles of H2 reacted = 35.7 mol * (3 mol H2 / 1 mol N2) = 107.1 mol H2 To calculate the mass of unreacted H2, first find the moles of unreacted H2: Moles of unreacted H2 = 248 mol (initial) - 107.1 mol (reacted) = 140.9 mol Finally, convert moles of unreacted H2 to mass: Mass of unreacted H2 = 140.9 mol * 2.016 g/mol = 2.84 x 10^2 g a. The maximum mass of ammonia that can be produced is \(1.22 \times 10^{3} g\). b. The starting material that would remain unreacted is hydrogen, with a mass of \(2.84 \times 10^{2} g\).

Key concepts

Understanding the Chemical Reaction Equation

The chemical reaction equation is at the heart of stoichiometry, which is the study of the quantitative relationships in a chemical reaction. In our example, the equation \( \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \) represents the synthesis of ammonia (NH3) from nitrogen (N2) and hydrogen (H2).

Each molecule and its quantity in the equation has a specific meaning: one molecule of N2 reacts with three molecules of H2 to produce two molecules of NH3. These ratios, known as stoichiometric coefficients, are crucial when predicting how much product can be formed from given amounts of reactants, or what amount of a particular reactant is necessary for the reaction to proceed.

The Role of the Limiting Reagent

In chemical reactions, the limiting reagent is the substance that is entirely consumed first and thus determines the amount of product formed. Identifying the limiting reagent is key to calculating theoretical yields. In the given problem, by comparing the amount of NH3 that could be produced from the available N2 and H2, we determined nitrogen to be the limiting reagent as it would produce less ammonia than the hydrogen would be able to.

This concept is especially important because it helps us realize that even if we have an excess of one reactant (in this case, hydrogen), the reaction cannot proceed once the limiting reagent (nitrogen) is used up. Thus, in practical applications, the cost-efficient operation of chemical processes often depends on optimizing the use of the limiting reagent.

How Molar Mass Connects Moles to Grams

Molar mass is a critical factor that bridges the gap between the microscopic scale (atoms and molecules) and the macroscopic world (grams). It is the mass of one mole of a substance and is expressed in grams per mole (g/mol). To solve stoichiometry problems, you must be able to convert between grams and moles using the molar mass.

In our exercise, we used the molar mass of nitrogen and hydrogen to convert the given masses to moles. This conversion is the first step in any stoichiometric calculation because chemical reaction equations reference moles, not grams. Knowing that the molar mass of N2 is 28.014 g/mol, we can understand why we need about 28 grams of N2 to have a mole of it, thereby giving us a base to calculate how much product we can expect from a given mass of reactants.

Moles to Grams Conversion: The Final Calculation

The moles to grams conversion is the final step in deriving the expected mass of a product from a stoichiometric calculation. After determining the moles of a substance produced in a reaction, multiply this quantity by the molar mass to find the corresponding mass in grams.

For instance, we found the number of moles of NH3 that could be produced from the limiting reagent (N2) and then multiplied it by the molar mass of NH3 to determine the maximum mass of ammonia produced. Conversely, to find the mass of remaining reactants, we subtract the moles of reactant that reacted, then convert the remaining moles back to grams. This conversion empowers students to visualize and measure the actual amount of materials involved in a chemical process.