Problem 68 - Introductory Chemistry Atoms First

Question

Consider the reaction \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)\) If this reaction takes place at STP, what volume in liters of \(\mathrm{H}_{2}\) gas would be required to react completely with \(22.4 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas? What volume in liters of \(\mathrm{H}_{2} \mathrm{O}(g)\) would be produced? Explain how you arrived at your answers.

Step-by-Step Solution

Answer

The volume of hydrogen gas required to react completely with 22.4 L of oxygen gas at STP is 44.8 L. The volume of water vapor produced from the reaction is also 44.8 L.

Step 1: Write down the balanced chemical equation

The balanced chemical equation for this reaction is: \[2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2}\mathrm{O}(g)\]

Step 2: Identify the molar volume of a gas at STP

At STP, which is defined as a temperature of 273.15K (0°C) and a pressure of 1 atm, 1 mole of any gas occupies 22.4 liters of volume. This is known as the molar volume of a gas at STP.

Step 3: Convert the volume of oxygen gas to moles

We are given that 22.4 L of O₂ gas is involved in the reaction. Using the molar volume of a gas at STP (22.4 L/mol), we can convert the volume of O₂ to moles: \[22.4 \,\text{L O}_2 * \frac{1 \,\text{mol O}_2}{22.4 \,\text{L O}_2} = 1\, \text{mol O}_2\]

Step 4: Use stoichiometry to find moles of hydrogen gas and water vapor

From the balanced chemical equation in step 1, we see that 2 moles of H₂ react with 1 mole of O₂ to form 2 moles of H₂O. Since we have 1 mole of O₂, the stoichiometry is as follows: 1. Moles of H₂ needed: \(2 \,\text{mol H}_2 * \frac{1 \,\text{mol O}_2}{1\, \text{mol O}_2} = 2\, \text{mol H}_2\) 2. Moles of H₂O produced: \(2 \,\text{mol H}_2\mathrm{O} * \frac{1 \,\text{mol O}_2}{1\, \text{mol O}_2} = 2\, \text{mol H}_2\mathrm{O}\)

Step 5: Convert moles to volume at STP

Using the molar volume of a gas at STP (22.4 L/mol), we can now convert the moles of hydrogen gas and water vapor back to their respective volumes: 1. Volume of H₂ required: \(2\, \text{mol H}_2 * \frac{22.4\, \text{L H}_2}{1\, \text{mol H}_2} = 44.8\, \text{L H}_2\) 2. Volume of H₂O produced: \(2\, \text{mol H}_2\mathrm{O} * \frac{22.4\, \text{L H}_2\mathrm{O}}{1\, \text{mol H}_2\mathrm{O}} = 44.8\, \text{L H}_2\mathrm{O}\)

Step 6: Report the final answer

The volume of hydrogen gas required to react completely with 22.4 L of oxygen gas at STP is 44.8 L. The volume of water vapor produced from the reaction is also 44.8 L.

Key Concepts

Understanding Chemical Reactions

When we talk about chemical reactions, we're referring to the process where substances, known as reactants, are transformed into new substances called products. The reaction outlined in the exercise, where hydrogen gas reacts with oxygen gas to form water vapor, is a fundamental example of a combination reaction. It's essential to note that in any chemical reaction, the mass and the number of atoms must be conserved. This concept, known as the law of conservation of mass, leads us directly to the importance of balancing chemical equations to reflect this conservation.

Each substance involved in a reaction is quantified in terms of 'moles', which are units that measure the amount of a substance based on the number of particles, specifically atoms or molecules, present. Balancing the equation ensures that the number of moles of the reactants equals the number of moles of the products. This helps students predict the outcome of reactions quantitatively and is critical for calculations in stoichiometry.

Molar Volume at STP

The molar volume at STP is an incredibly useful concept when working with gases. At Standard Temperature and Pressure (0°C and 1 atmosphere of pressure), one mole of any ideal gas occupies 22.4 liters of space. This standard measurement allows chemists and students alike to easily convert between the volume of gas and the number of moles. By knowing this constant value, we can solve for unknown variables in a chemical reaction that takes place under these conditions. For instance, if we want to know the volume of a gas required or produced as in our exercise, we can use the balanced chemical equation in conjunction with the molar volume at STP to find our answer.

Remember, however, that real gases may deviate slightly from this volume due to intermolecular forces and the volume occupied by the gas molecules themselves, but for many cases, including homework problems, using 22.4 L/mol is sufficiently accurate.

Balancing Chemical Equations

The cornerstone of stoichiometry lies in balancing chemical equations. A balanced equation ensures that there's an equal number of each type of atom on both the reactant and the product sides. This reflects the law of conservation of mass and provides the basis for stoichiometric calculations. When balancing an equation, we adjust coefficients—the numbers in front of the chemical formulas—to ensure this balance.

Once an equation is balanced, these coefficients represent the ratio of moles that react and are formed, known as the stoichiometric ratio. For instance, in our exercise, the balanced equation has coefficients that show two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water vapor. It doesn't matter if you are working with milligrams or tons, or liters or gallons – the stoichiometric ratio remains the same and is crucial for scaling reactions to any desired amount.