Question

Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the equation: $$\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)$$ In a particular reaction, 25.5 \(\mathrm{L}\) of methane gas (at a pressure of 732 torr and a temperature of \(25^{\circ} \mathrm{C} )\) mixes with 22.8 \(\mathrm{L}\) of water vapor \((\mathrm{at}\) a pressure of 702 torr and a temperature of \(125^{\circ} \mathrm{C} )\) . The reaction produces 26.2 \(\mathrm{L}\) of hydrogen gas at STP. What is the percent yield of the reaction?

Short answer

The percent yield is the ratio of the actual volume of hydrogen gas produced at STP to the theoretical volume, multiplied by 100.

Step-by-step solution

Convert Volumes to Moles using Ideal Gas Law

First convert the given volumes of methane and water vapor to moles using the ideal gas law. For methane: \( PV = nRT \), solve for \( n \) using \( P = 732 \) torr, \( V = 25.5 \) L, \( R = 62.36 \) L torr K\textsuperscript{-1} mol\textsuperscript{-1}, and \( T = 25^{\text{\textdegree}}C = 298K \). For water vapor: use \( P = 702 \) torr, \( V = 22.8 \) L, and \( T = 125^{\text{\textdegree}}C = 398K \).

Determine the Limiting Reactant

Compare the mole ratio of methane to water vapor with the stoichiometry of the reaction to find the limiting reactant. The reaction shows a 1:1 mole ratio, so whichever has fewer moles will be the limiting reactant.

Calculate Theoretical Yield in Moles of Hydrogen Gas

Based on the limiting reactant, calculate the theoretical yield of hydrogen gas in moles using the 1:3 stoichiometric ratio from the reaction equation.

Convert Theoretical Yield to Liters at STP

Using the ideal gas law, convert the moles of theoretical hydrogen gas to volume at STP (Standard Temperature and Pressure: \(0^{\text{\textdegree}}C\), \(1 \) atm). Use \( V = nRT/P \) with \( R = 0.0821 \) L atm K\textsuperscript{-1} mol\textsuperscript{-1}, \( T = 273K \), and \( P = 1 \) atm.

Calculate Percent Yield

The percent yield is the actual yield (the given 26.2 L of \( H_{2} \) at STP) divided by the theoretical yield (from Step 4) multiplied by 100. Percent yield = \( \frac{actual \text{ yield }}{theoretical \text{ yield }} \times 100\% \).

Key concepts

Ideal Gas Law

Understanding the ideal gas law is crucial when dealing with gases in chemical reactions. It is represented by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. This equation allows us to calculate the amount of substance in moles when the pressure, volume, and temperature of a gas are known.

In the context of the given exercise, the ideal gas law is used to convert the volumes of methane and water vapor to moles before reacting them. Since the pressures and temperatures differ for the two gases, using the ideal gas law ensures accurate stoichiometric calculations. It is important to convert the temperatures to Kelvin and use the appropriate value for R based on the units of pressure provided in the question.

Limiting Reactant

The limiting reactant is the substance in a chemical reaction that is completely consumed first, thus determining the maximum amount of product that can be formed. It is essential to identify the limiting reactant because it dictates the theoretical yield—the maximum amount of product expected based on stoichiometry. Identifying the limiting reactant requires a comparison of the mole ratios of reactants to the ratios indicated in the balanced chemical equation.

For the exercise, by comparing the mole ratio of methane to water vapor to the stoichiometric ratio from the balanced equation, we can identify which reactant will limit the production of hydrogen gas, thereby allowing us to proceed with the calculation of theoretical and percent yields.

Stoichiometry

Stoichiometry is the area of chemistry that pertains to the quantitative relationships between reactants and products in a chemical reaction. It is based on the conserved nature of mass in chemical reactions, as dictated by the Law of Conservation of Mass. In practice, stoichiometry utilizes the mole ratios provided by balanced chemical equations to predict the amounts of substances consumed and produced.

In our problem, the balanced chemical equation shows that one mole of methane reacts with one mole of water vapor to produce one mole of carbon monoxide and three moles of hydrogen gas. These stoichiometric relationships are fundamental to calculating the theoretical yield of hydrogen gas based on the amount of limiting reactant.

Theoretical Yield

The theoretical yield of a chemical reaction is the maximum amount of product that can be generated as predicted by stoichiometry, assuming that the reaction goes to completion and all of the limiting reactant is converted into the product. To calculate the theoretical yield, one must use the mole ratio between the limiting reactant and the product from the balanced chemical equation.

In the provided exercise, by applying stoichiometry, we determine the theoretical yield of hydrogen gas in moles. This quantity is then converted to liters at STP using the ideal gas law to compare with the actual yield and calculate the percent yield of the reaction. The theoretical yield is vital for determining the efficiency of the reaction through the percent yield calculation, which is the actual yield divided by the theoretical yield, multiplied by 100%. This gives insight into any potential losses or inefficiencies in the reaction process.