Problem 78 - Introductory Chemistry Atoms First

Question

Consider the following balanced chemical equation: \(2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) (a) How many grams of water are formed from \(5.00 \mathrm{~g}\) of \(\mathrm{H}_{2}\) and an excess amount of \(\mathrm{O}_{2}\) ? (b) How many grams of \(\mathrm{O}_{2}\) do you need to produce \(5.00 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) ? (c) Given \(100.0 \mathrm{~g}\) of \(\mathrm{H}_{2}\), how many grams of \(\mathrm{O}_{2}\) are required to run the reaction in a stoichiometric fashion? (d) What is the theoretical yield in grams of water upon combining \(50.0 \mathrm{~g}\) of \(\mathrm{O}_{2}\) with an excess amount of \(\mathrm{H}_{2}\) ? (e) Express the answer to part (d) in terms of the number of water molecules.

Step-by-Step Solution

Answer

(a) 44.687 g of water are formed from 5.00 g of H2 and an excess amount of O2. (b) 4.4392 g of O2 are needed to produce 5.00 g of H2O. (c) 793.728 g of O2 are required to react with 100.0 g of H2 in a stoichiometric fashion. (d) The theoretical yield in grams of water upon combining 50.0 g of O2 with an excess amount of H2 is 56.298 g. (e) There are approximately \(1.881 \times 10^{24}\) water molecules in the product.

Step 1: (a) Calculate moles of H2

First, we need to determine the moles of hydrogen gas, H2, given in the question, which is 5.00 g. We know that the molar mass of H2 is approximately 2.016 g/mol, thus we have: moles of H2 = (5.00 g) / (2.016 g/mol) = 2.4802 mol

Step 2: (a) Calculate moles of H2O

Since the balanced chemical equation states that 2 moles of H2 yields 2 moles of H2O, we can say that the mol ratio is 1:1, so: moles of H2O = moles of H2 = 2.4802 mol

Step 3: (a) Calculate grams of H2O

Now, we need to find the mass of H2O formed by this reaction. We know the molar mass of H2O is 18.015 g/mol. Hence, grams of H2O = moles of H2O * molar mass of H2O = 2.4802 mol * 18.015 g/mol = 44.687 g Thus, 44.687 g of water are formed from 5.00 g of H2 and an excess amount of O2.

Step 4: (b) Calculate moles of H2O

In this scenario, we are given 5.00 g of H2O and need to calculate the moles of H2O. We know the molar mass of H2O is 18.015 g/mol, so: moles of H2O = (5.00 g) / (18.015 g/mol) = 0.2775 mol

Step 5: (b) Calculate moles of O2

From the balanced chemical equation, we know that 1 mole of O2 reacts with 2 moles of H2 to produce 2 moles of H2O. Thus, the ratio of moles between O2 and H2O is 1:2. So, the moles of O2 needed will be: moles of O2 = moles of H2O / 2 = 0.2775 mol / 2 = 0.1387 mol

Step 6: (b) Calculate grams of O2

Now, we need to find the mass of O2 needed to produce 5.00 g of H2O. The molar mass of O2 is 32.00 g/mol, so: grams of O2 = moles of O2 * molar mass of O2 = 0.1387 mol * 32.00 g/mol = 4.4392 g Thus, 4.4392 g of O2 are needed to produce 5.00 g of H2O.

Step 7: (c) Calculate moles of H2

Given 100.0 g of H2, we need to determine the moles of H2. We know that the molar mass of H2 is approximately 2.016 g/mol, so: moles of H2 = (100.0 g) / (2.016 g/mol) = 49.608 mol

Step 8: (c) Calculate moles of O2

From the balanced chemical equation, we know that 2 moles of H2 reacts with 1 mole of O2 to produce 2 moles of H2O. So, the moles of O2 needed will be: moles of O2 = moles of H2 / 2 = 49.608 mol / 2 = 24.804 mol

Step 9: (c) Calculate grams of O2

Now, we need to find the mass of O2 needed to react with 100.0 g of H2. The molar mass of O2 is 32.00 g/mol, so: grams of O2 = moles of O2 * molar mass of O2 = 24.804 mol * 32.00 g/mol = 793.728 g Thus, 793.728 g of O2 are required to react with 100.0 g of H2 in a stoichiometric fashion.

Step 10: (d) Calculate moles of O2

Given 50.0 g of O2, we need to determine the moles of O2. We know that the molar mass of O2 is 32.00 g/mol, so: moles of O2 = (50.0 g) / (32.00 g/mol) = 1.5625 mol

Step 11: (d) Calculate moles of H2O

From the balanced chemical equation, we know that 1 mole of O2 reacts with 2 moles of H2 to produce 2 moles of H2O. Thus, the ratio of moles between O2 and H2O is 1:2. So, the moles of H2O produced will be: moles of H2O = moles of O2 * 2 = 1.5625 mol * 2 = 3.125 mol

Step 12: (d) Calculate grams of H2O

Now, we need to find the mass of H2O formed by this reaction. We know the molar mass of H2O is 18.015 g/mol, so: grams of H2O = moles of H2O * molar mass of H2O = 3.125 mol * 18.015 g/mol = 56.298 g Thus, the theoretical yield in grams of water upon combining 50.0 g of O2 with an excess amount of H2 is 56.298 g.

Step 13: (e) Calculate number of water molecules

In this step, we need to convert moles of H2O to the number of water molecules. We know that 1 mole contains Avogadro's number of particles (about \(6.022 \times 10^{23}\) particles), so: number of water molecules = moles of H2O * Avogadro's Number = 3.125 mol * \(6.022 \times 10^{23}\) molecules/mol ≈ \(1.881 \times 10^{24}\) molecules Thus, there are approximately \(1.881 \times 10^{24}\) water molecules in the product.

Key Concepts

Chemical Reaction Balancing

Understanding how to balance chemical reactions is fundamental in chemistry. It ensures that the same number of atoms for each element is present on both sides of the equation, complying with the law of conservation of mass. For instance, the balanced reaction \(2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}\) shows the exact ratio of hydrogen (\(H_2\)) and oxygen (\(O_2\)) molecules needed to produce water (\(H_2O\)). Maintaining balanced equations allows for precise stoichiometric calculations.

Importance in Calculations

Balancing chemical reactions is a precursor to stoichiometric calculations. It informs the mole ratio, which is a conversion factor between substances in a reaction. In the given problem, the balanced equation indicates that 2 moles of hydrogen gas react with 1 mole of oxygen gas to form 2 moles of water—this 1:1 mole ratio between \(H_2\) and \(H_2O\) is crucial in determining the amounts of reactants and products.

Molar Mass

Molar mass is a property that links the mass of a substance to its amount in moles. It is expressed in grams per mole (\(\text{g/mol}\)) and is equivalent to the mass of one mole of a substance. For example, the molar mass of hydrogen gas (\(H_2\)) is approximately 2.016 \(\text{g/mol}\), while the molar mass of water (\(H_2O\)) is 18.015 \(\text{g/mol}\).

Calculation Purpose

Molar mass is essential for converting between the mass of a substance and the number of moles, as shown in the steps provided in the solution. By using the molar mass, you can determine the mass of water formed or the mass of a reactant needed from the respective number of moles.

Theoretical Yield

The concept of theoretical yield is a critical aspect of stoichiometry and chemistry as a whole. It represents the maximum amount of product that could be formed from a given amount of reactants, according to the stoichiometry of the chemical reaction. It is purely theoretical because it assumes perfect efficiency without any losses.

Real-life Application

In the given problem, determining the theoretical yield helps us understand how much water can be produced when \(50.0 \text{g}\) of oxygen is reacted with an excess of hydrogen gas. This process involves calculating the number of moles of reactants, using the mole ratio from the balanced equation, and finally applying the molar masses of the products to get the theoretical yield in grams.

Avogadro's Number

Avogadro's number is a fundamental constant in chemistry that is used to express the number of particles, usually atoms or molecules, in one mole of a substance. It is approximately \(6.022 \times 10^{23}\). This large number bridges the gap between the atomic scale and the macroscopic scale we interact with in the laboratory.

Conversion to Particles

In the context of the textbook problem, Avogadro's number enables us to convert moles of water to the actual number of water molecules—the final form in which students can visualize the amount of product formed. For instance, the produced \(3.125 \text{moles of } H_2O\) are converted to molecules using Avogadro's number, yielding \(1.881 \times 10^{24}\) molecules of water, illustrating the vast number of particles involved in chemical reactions.