Problem 95 - Chemistry An Atoms First Approach

Question

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be produced by the following reaction: $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)$$ Hydrogen at STP flows into a reactor at a rate of \(16.0 \mathrm{L} / \mathrm{min.}\) Carbon monoxide at STP flows into the reactor at a rate of \(25.0 \mathrm{L} / \mathrm{min.}\). If \(5.30\) \(\mathrm{g}\) methanol is produced per minute, what is the percent yield of the reaction?

Step-by-Step Solution

Answer

The percent yield for the given reaction can be calculated using the following steps: 1. Calculate moles of CO and H₂ using the given volumes and STP conditions. 2. Determine the limiting reactant by comparing mole ratios. 3. Calculate the theoretical mass of Methanol based on the limiting reactant. 4. Calculate the percent yield using the formula: \(\text{Percent Yield} = \frac{\text{Actual Mass of CH}_{3}\text{OH}}{\text{Theoretical Mass of CH}_{3}\text{OH}} * 100\) Plug in the values and calculate the percent yield.

Step 1: Calculate moles of the given reactants

To calculate the theoretical Methanol production, we first need to convert the given volumes of CO and H₂ at STP into moles. We can use the equation: n = PV/RT At STP (Standard Temperature and Pressure), R = 0.0821 L atm K⁻¹ mol⁻¹, T = 273.15 K, and P = 1 atm. Thus, equation becomes: n = V/R T For the given volumes of CO and H₂, plug into the equation and calculate their moles: Moles of CO = \(25.0 / (0.0821 * 273.15)\) Moles of H₂ = \(16.0 / (0.0821 * 273.15)\)

Step 2: Determine the limiting reactant

Compare the mole ratio of CO and H₂ according to the balanced equation (ratio is 1:2). Then, calculate the limiting reactant and how many moles of Methanol can be produced as per the stoichiometric ratio. For the given moles of CO and H₂, calculate the ratios: Ratio_CO = Moles of CO Ratio_H₂ = Moles of H₂ / 2 Compare these ratios and determine the limiting reactant and calculate the number of moles of Methanol that can be theoretically produced: If Ratio_CO < Ratio_H₂, then CO is the limiting reactant and moles of CH₃OH = moles of CO If Ratio_H₂ < Ratio_CO, then H₂ is the limiting reactant and moles of CH₃OH = moles of H₂ / 2

Step 3: Calculate the theoretical mass of Methanol

Now, convert the moles of Methanol we determined based on limiting reactant to mass using molecular weight (32 g/mol). Mass of CH₃OH (theoretical) = Moles of CH₃OH * 32 g/mol

Step 4: Calculate the percent yield

In the end, calculate the percent yield using the formula: Percent Yield = \(\frac{Actual \ Mass \ of \ CH_{3}OH}{Theoretical \ Mass \ of \ CH_{3}OH} * 100\) Percentage yield = \(\frac{5.30}{Mass \ of \ CH_{3}OH \ (theoretical)} * 100\) This will give us the percent yield for the given reaction.

Key Concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that gets completely used up first. This reactant limits the amount of product that can be formed. Think of it like making sandwiches: if you have bread and cheese but run out of bread first, bread is the limiting element.
For the methanol production, it's essential to identify which reactant limits the reaction. We use the balanced chemical equation:

• CO + 2H₂ → CH₃OH

Using this, we can compare the moles of each reactant. This helps us calculate how much of each ingredient we have to work with.
By determining the limiting reactant, we know how much methanol can theoretically be produced. It's all about the smallest amount needed according to the ratio!

Stoichiometry

Stoichiometry is the study of reactant and product quantities in chemical reactions. It tells us how much of each reactant is needed and how much product is formed.
Let's consider our balanced equation:

• 1 mole of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH

This is a classic stoichiometric ratio.
Understanding stoichiometry allows chemists to predict product quantities and ensures that they use just the right amounts of ingredients. This is especially useful in industrial applications where efficiency is crucial.

Theoretical Yield

Theoretical yield is the maximum amount of product that a reaction can generate from the given quantities of reactants.
It's calculated based on the limiting reactant, as this dictates how far the reaction can go.

• In our methanol example, once we identify the limiting reactant, we use it to find how many moles of methanol we should theoretically get.
• This involves using the stoichiometric ratios provided by the balanced chemical equation.

By converting these moles into grams, using the molecular weight of methanol (32 g/mol), we obtain the theoretical mass of our product.

Moles Calculation

Moles are a fundamental concept in chemistry, representing a specific number of particles. In chemical calculations, we often convert between volume, mass, and moles.
At STP (Standard Temperature and Pressure), we use the formula:

• n = V / (R * T)

where n is the number of moles, V is volume, R is the gas constant, and T is the temperature.
This allows us to find how many moles of CO and H₂ are entering the reactor. Once we have the moles, we can easily move forward with stoichiometry and limiting reactant calculations.

• This enables accurate determination of product yield in reactions such as our methanol production.