Problem 73 - Chemistry: Principles and Reactions

Question

A gaseous mixture containing \(4.15 \mathrm{~mol}\) of hydrogen gas and \(7.13 \mathrm{~mol}\) of oxygen gas reacts to form steam. (a) Write a balanced equation for the reaction. (b) What is the limiting reactant? (c) What is the theoretical yield of steam in moles? (d) How many moles of the excess reactant remain unreacted?

Step-by-Step Solution

Answer

Question: In a gaseous mixture, 4.15 moles of hydrogen gas and 7.13 moles of oxygen gas react to form steam. Identify the limiting reactant, the theoretical yield of steam in moles, and how many moles of excess reactant remain unreacted. Answer: The limiting reactant is hydrogen gas. The theoretical yield of steam is 4.15 moles. There are 5.055 moles of excess oxygen reactant remaining unreacted.

Step 1: Write the balanced equation for the reaction

The balanced chemical equation for the reaction between hydrogen and oxygen to form water (steam) is: $$2\mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(g)$$

Step 2: Determine the limiting reactant

To determine the limiting reactant, we need to compare the mole ratios of the reactants. The balanced equation tells us that for every \(2\,\text{mol}\) of \(\mathrm{H}_{2}\), we need \(1\,\text{mol}\) of \(\mathrm{O}_{2}\). For the given moles of reactants: Hydrogen gas: \(4.15 \,\text{mol}\) Oxygen gas: \(7.13 \,\text{mol}\) Calculate the mole ratios for each reactant: Hydrogen gas: \(\frac{4.15}{2} = 2.075\) Oxygen gas: \(\frac{7.13}{1} = 7.13\) Since \(2.075\) is smaller than \(7.13\), hydrogen gas is the limiting reactant.

Step 3: Calculate the theoretical yield of steam

Using the limiting reactant, we can find the theoretical yield of steam. From the balanced equation, we know that \(2\,\text{mol}\) of \(\mathrm{H}_{2}\) produce \(2\,\text{mol}\) of \(\mathrm{H}_{2}\mathrm{O}\). We have \(4.15\,\text{mol}\) of \(\mathrm{H}_{2}\): Theoretical yield of steam: \((4.15\,\text{mol}\,\mathrm{H}_{2}) \times \frac{2\,\text{mol}\,\mathrm{H}_{2}\mathrm{O}}{2\,\text{mol}\,\mathrm{H}_{2}} = 4.15\,\text{mol}\,\mathrm{H}_{2}\mathrm{O}\) So, the theoretical yield of steam is \(4.15\,\text{mol}\).

Step 4: Calculate the moles of excess reactant that remain unreacted

To find the moles of unreacted oxygen, we will use the ratio of the limiting reactant to the excess reactant from the balanced equation. The ratio is \(\frac{2\,\text{mol}\,\mathrm{H}_{2}}{1\,\text{mol}\,\mathrm{O}_{2}}\). We know that \(4.15\,\text{mol}\) of hydrogen gas reacted completely: Moles of reacted oxygen: \((4.15\,\text{mol}\,\mathrm{H}_{2}) \times \frac{1\,\text{mol}\,\mathrm{O}_{2}}{2\,\text{mol}\,\mathrm{H}_{2}} = 2.075\,\text{mol}\,\mathrm{O}_{2}\) Moles of unreacted oxygen: \(7.13\,\text{mol}\,\mathrm{O}_{2} - 2.075\,\text{mol}\,\mathrm{O}_{2} = 5.055\,\text{mol}\,\mathrm{O}_{2}\) So, there are \(5.055\,\text{mol}\) of excess oxygen reactant remaining unreacted.

Key Concepts

Theoretical Yield Calculation

Understanding how to calculate theoretical yield is essential for students delving into the world of chemical reactions. It represents the maximum amount of product that can be formed in a chemical reaction, based solely on the amount of limiting reactant.

Let's apply this to an exercise example where hydrogen and oxygen react to form water. After determining the limiting reactant, hydrogen, with 4.15 moles available, we can ascertain the theoretical yield. Since the balanced equation shows a 1:1 mole ratio between the reactant hydrogen and the product water, simply put, the theoretical yield in moles of water will match that of the reactant hydrogen, which is 4.15 moles.

An important aspect to remember is to always use the limiting reactant to calculate the theoretical yield. Doing otherwise would result in an incorrect yield that cannot physically be obtained due to insufficient reactant quantities.

Chemical Reaction Stoichiometry

Stoichiometry is the cornerstone of analytical chemistry that involves calculating the quantities of reactants and products in chemical reactions. It's all about the numbers and their ratios. The central principle at play is the law of conservation of mass, where the mass of the reactants equals the mass of the products in a closed system.

Using the stoichiometry of the balanced equation which in our exercise is \(2\mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(g)\), we compare the amount of available reactants with the ratios indicated by the equation. By doing so, we can pinpoint the limiting reactant, determine the theoretical yield, and find out how much of the excess reactant remains after the reaction. Students must become familiar with using stoichiometric coefficients to relate the quantities of reactants and products accurately.

Balanced Chemical Equations

At the heart of each chemical reaction lies a balanced chemical equation, which ensures that atoms are conserved throughout the process, aligning with the law of conservation of mass. A balanced equation has equal numbers of each type of atom on both sides of the reaction.

For instance, the reaction between hydrogen and oxygen to create water is depicted by the equation we've discussed: \(2\mathrm{H}_2(g) + \mathrm{O}_2(g) \rightarrow 2\mathrm{H}_2\mathrm{O}(g)\). Note the stoichiometric coefficients: '2' in front of hydrogen and water, indicating that two moles of hydrogen combine with one mole of oxygen to produce two moles of water. Balancing equations is a critical step that informs not only the stoichiometry calculations but also the identification of the limiting reactant and the computation of the theoretical yield.

Remember that sometimes, balancing chemical equations may involve coefficients that are fractions. In such cases, multiply through by the smallest number to convert all coefficients to whole numbers, simplifying calculations and interpretation.