Problem 80 - Introductory Chemistry Atoms First

Question

Consider the following unbalanced chemical equation: \(\mathrm{H}_{2}+\mathrm{N}_{2} \rightarrow \mathrm{NH}_{3}\) (a) To run this reaction in a balanced fashion, how much nitrogen is required if you start with \(10.0 \mathrm{~g}\) of \(\mathrm{H}_{2} ?\) (b) How many grams of ammonia \(\left(\mathrm{NH}_{3}\right)\) will you produce if you run the reaction with the masses calculated in part (a)? (c) How many molecules of ammonia will you produce?

Step-by-Step Solution

Answer

(a) 46.3 grams of nitrogen are needed for a balanced reaction with 10.0 grams of hydrogen. (b) 56.3 grams of ammonia will be produced when the reaction runs with the masses calculated in part (a). (c) \(1.989 \times 10^{24}\) molecules of ammonia will be produced.

Step 1: Balance the chemical equation

To balance the chemical equation, we need to make sure that the number of atoms for each element on the reactant side is equal to the number of atoms on the product side. The balanced chemical equation is: \[3\mathrm{H}_{2}+\mathrm{N}_{2} \rightarrow 2\mathrm{NH}_{3}\]

Step 2: Calculate the moles of hydrogen and find the moles of nitrogen needed

First, we need to convert the given mass of hydrogen into moles. Using the molar mass of hydrogen (1.008 g/mol): \[10.0 \text{ g } \mathrm{H}_{2} \times \frac{1 \text{ mol }}{2(1.008 \text{ g})} = \frac{10.0}{2.016} = 4.960 \text{ mol }\mathrm{H}_{2}\] Using the stoichiometry from the balanced chemical equation, it takes 3 moles of hydrogen to react with 1 mole of nitrogen to produce 2 moles of ammonia. Hence the number of moles of nitrogen required is: \[4.960 \text{ mol }\mathrm{H}_{2} \times \frac{1 \text{ mol }\mathrm{N}_{2}}{3 \text{ mol }\mathrm{H}_{2}} = \frac{4.960}{3} = 1.653 \text{ mol }\mathrm{N}_{2}\]

Step 3: Calculate the mass of nitrogen required

Now that we have found the moles of nitrogen needed, let's convert it to grams. We can obtain the mass using the molar mass of nitrogen (14.007 g/mol): \[1.653 \text{ mol }\mathrm{N}_{2} \times \frac{28.014 \text{ g }}{1 \text{ mol }\mathrm{N}_{2}} = 1.653 \times 28.014 = 46.3 \text{ g }\mathrm{N}_{2}\] So, 46.3 grams of nitrogen are needed for a balanced reaction with 10.0 grams of hydrogen.

Step 4: Determine the mass of ammonia produced

Using the stoichiometry of the balanced chemical equation, we can find the mass of the ammonia produced. For this, we can use the moles of hydrogen calculated in Step 2. \[4.960 \text{ mol }\mathrm{H}_{2} \times \frac{2 \text{ mol }\mathrm{NH}_{3}}{3 \text{ mol }\mathrm{H}_{2}} = \frac{2(4.960)}{3} = 3.307 \text{ mol }\mathrm{NH}_{3}\] Let's convert the moles of ammonia into grams: \[3.307 \text{ mol }\mathrm{NH}_{3} \times \frac{17.031 \text{ g }}{1 \text{ mol }\mathrm{NH}_{3}} = 3.307 \times 17.031 = 56.3 \text{ g }\mathrm{NH}_{3}\] So, when the reaction runs with the masses calculated in part (a), 56.3 grams of ammonia will be produced.

Step 5: Calculate the number of ammonia molecules produced

To find the number of molecules of ammonia, we can use Avogadro's number (approximately \(6.022 \times 10^{23}\)): \[3.307 \text{ mol }\mathrm{NH}_{3} \times \frac{6.022 \times 10^{23} \text{ molecules}}{1 \text{ mol }} = 3.307 \times 6.022 \times 10^{23} = 1.989 \times 10^{24} \text{ molecules }\mathrm{NH}_{3}\] So, 1.989 x \(10^{24}\) molecules of ammonia will be produced in this reaction.

Key Concepts

Stoichiometry

Stoichiometry is the aspect of chemistry that pertains to the calculation of the quantities of reactants and products in chemical reactions. It's based on the law of conservation of mass that states that matter is not created nor destroyed in a chemical reaction. This means that the mass and number of atoms of the reactants equal the mass and number of atoms of the products.

In practical terms, stoichiometry involves using the coefficients of a balanced chemical equation to derive the ratio between the amounts of each substance involved. For example, in the balanced reaction \(3\mathrm{H}_{2} + \mathrm{N}_{2} \rightarrow 2\mathrm{NH}_{3}\), stoichiometry tells us that three moles of hydrogen gas will react with one mole of nitrogen gas to produce two moles of ammonia. Therefore, if you have a certain amount of one reactant (say hydrogen), you can calculate how much of the other reactant (nitrogen) is required for the reaction to proceed without any leftover hydrogen. Similarly, stoichiometry can also be used to predict the amount of product formed from given quantities of reactants.

Molar Mass Calculations

Molar mass is a fundamental concept in chemistry that represents the mass of one mole of a substance. One mole contains exactly \(6.022 \times 10^{23}\) entities (atoms, molecules, ions, etc.) of that substance. The molar mass is typically expressed in grams per mole (g/mol) and is calculated by summing the atomic masses of all the atoms in a molecule.

For instance, to calculate the molar mass of water (\(\mathrm{H}_{2}O\)), you'd add together twice the atomic mass of hydrogen (approximately 1.008 g/mol) and the atomic mass of oxygen (approximately 16.00 g/mol), giving us approximately 18.016 g/mol for water.

Molar mass calculations are central to converting between the mass of a substance and the amount in moles, which allows for the use of stoichiometry in quantitative chemical analysis. This process was utilized in the original exercise to determine the masses required for a balanced reaction and the mass of ammonia produced from a given amount of hydrogen.

Avogadro's Number

Avogadro's number, approximately \(6.022 \times 10^{23}\), is a constant that represents the number of atoms, molecules, or ions in one mole of a substance. It's named after Amedeo Avogadro, an Italian scientist who first proposed that the volume of a gas (at a given pressure and temperature) is proportional to the number of atoms or molecules, regardless of the type of gas.

Thanks to Avogadro's number, chemists can relate the macroscopic world that we can measure (like mass in grams) to the microscopic world of atoms and molecules. For instance, if you have one mole of a substance, you know that you have \(6.022 \times 10^{23}\) units of that substance. This relation was applied in the original exercise to calculate the number of ammonia molecules produced in the reaction. Understanding Avogadro's number is critical when working with chemical equations and performing calculations that involve counting individual atoms or molecules.