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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?

Short Answer

Expert verified
The percent yield for the given reaction can be calculated using the following steps: 1. Calculate moles of CO and H鈧 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 by step solution

01

Calculate moles of the given reactants

To calculate the theoretical Methanol production, we first need to convert the given volumes of CO and H鈧 at STP into moles. We can use the equation: n = PV/RT At STP (Standard Temperature and Pressure), R = 0.0821 L atm K鈦宦 mol鈦宦, T = 273.15 K, and P = 1 atm. Thus, equation becomes: n = V/R T For the given volumes of CO and H鈧, plug into the equation and calculate their moles: Moles of CO = \(25.0 / (0.0821 * 273.15)\) Moles of H鈧 = \(16.0 / (0.0821 * 273.15)\)
02

Determine the limiting reactant

Compare the mole ratio of CO and H鈧 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 H鈧, calculate the ratios: Ratio_CO = Moles of CO Ratio_H鈧 = Moles of H鈧 / 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_H鈧, then CO is the limiting reactant and moles of CH鈧僌H = moles of CO If Ratio_H鈧 < Ratio_CO, then H鈧 is the limiting reactant and moles of CH鈧僌H = moles of H鈧 / 2
03

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 CH鈧僌H (theoretical) = Moles of CH鈧僌H * 32 g/mol
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 + 2H鈧 鈫 CH鈧僌H
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 H鈧 to produce 1 mole of CH鈧僌H
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 H鈧 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.

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Most popular questions from this chapter

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