Question

Methanol (CH_l3 \(\mathrm{OH}\) ) 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 L/min. Carbon monoxide at STP flows into the reactor at a rate of 25.0 L/min. If 5.30 g methanol is produced per minute, what is the percent yield of the reaction?

Short answer

The percent yield of methanol in this reaction is 46.3%.

Step-by-step solution

Convert L/min flow rates of reactants into moles/min

At STP (standard temperature and pressure), 1 mole of an ideal gas occupies 22.4 L. We can use this information to convert the flow rates of hydrogen and carbon monoxide into moles/min: For hydrogen: \(\frac{16.0\,\text{L}}{\text{min}}\cdot\frac{1\,\text{mol}}{22.4\,\text{L}}=0.714\,\text{mol}\,\text{H}_{2}\,\text{min}^{-1}\) For carbon monoxide: \(\frac{25.0\,\text{L}}{\text{min}}\cdot\frac{1\,\text{mol}}{22.4\,\text{L}}=1.12\,\text{mol}\,\text{CO}\,\text{min}^{-1}\)

Determine the limiting reactant

To determine the limiting reactant, first find the ratio between the actual flow rates in mol/min and the stoichiometric coefficients from the balanced equation: For hydrogen: \(\frac{0.714\,\text{mol}\,\text{H}_{2}\,\text{min}^{-1}}{2}=0.357\) For carbon monoxide: \(\frac{1.12\,\text{mol}\,\text{CO}\,\text{min}^{-1}}{1}=1.12\) The lower ratio belongs to hydrogen, so it is the limiting reactant.

Calculate the theoretical yield of methanol

Now, we can use stoichiometry to determine the theoretical yield of methanol (CH₃OH). According to the balanced equation, 1 mol CO reacts with 2 mol H₂ to produce 1 mol CH₃OH. As hydrogen is the limiting reactant, we have: Theoretical yield of methanol = \(0.714\,\text{mol}\,\text{H}_{2}\,\text{min}^{-1}\cdot\frac{1\,\text{mol}\,\text{CH}_{3}\text{OH}}{2\,\text{mol}\,\text{H}_{2}}=0.357\,\text{mol}\,\text{CH}_{3}\text{OH}\,\text{min}^{-1}\) We need to convert this amount into grams: Theoretical yield of methanol = \(0.357\,\text{mol}\,\text{CH}_{3}\text{OH}\,\text{min}^{-1}\cdot\frac{32.04\,\text{g}}{1\,\text{mol}}=11.43\,\text{g}\,\text{CH}_{3}\text{OH}\,\text{min}^{-1}\)

Calculate the percent yield

Now, we can use the actual yield (5.30 g/min) and the theoretical yield (11.43 g/min) to calculate the percent yield: Percent yield = \(\frac{5.30\,\text{g}\,\text{CH}_{3}\text{OH}\,\text{min}^{-1}}{11.43\,\text{g}\,\text{CH}_{3}\text{OH}\,\text{min}^{-1}}\times100\%=46.3\%\) So the percent yield of methanol in this reaction is 46.3%.

Key concepts

Stoichiometry

Stoichiometry is like a recipe in chemistry. It tells you how much of each chemical you will need and produce in a reaction. Basically, it is the calculation of reactants and products in chemical reactions. This is vital because it helps us to predict the quantities of substances consumed and produced.
To perform stoichiometry calculations, start with a balanced chemical equation. This step ensures we have the exact proportions of elements needed for the reaction. Next, use the coefficients from the balanced equation to create ratios. These ratios serve as conversion factors that guide us through the calculations.
For methanol production, the balanced equation is:\[ \text{CO}(g) + 2 \ \text{H}_{2}(g) \rightarrow \text{CH}_{3}\text{OH}(g) \]From this, we know that 1 mole of carbon monoxide will react with 2 moles of hydrogen to produce 1 mole of methanol. Knowing these relationships allows us to calculate the theoretical yield, which is what should be produced in an ideal situation.

Limiting Reactant

In many reactions, one reactant will run out before the others; this is the limiting reactant. It determines the maximum amount of product that can be made. If you think of it like baking cookies, the limiting reactant is the ingredient you run out of first. Even if you have plenty of other ingredients, you can't make more cookies without the limiting one.
To identify the limiting reactant, you must compare the ratio of reactants available against the stoichiometric ratio from the balanced equation.

• First, convert the amounts of reactants to moles.
• Next, divide these actual moles by their respective coefficients from the balanced equation.

In the given reaction scenario, hydrogen gas acted as the limiting reactant. Since it has the smaller ratio when compared, it determines the extent of the reaction and the theoretical yield of methanol.

Ideal Gas Law

The ideal gas law connects the four properties of gases: pressure, volume, temperature, and number of moles. It's expressed with the equation \( PV = nRT \), where:

• \( P \) is the pressure of the gas
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the ideal gas constant
• \( T \) is the temperature in Kelvin

In this reaction, since we are dealing with gases at STP (standard temperature and pressure), we use a simplified information that 1 mole of an ideal gas occupies 22.4 liters. This helps in converting the flow rates of carbon monoxide and hydrogen gas from volumes per minute into moles per minute. This conversion is crucial to determining the stoichiometry of the reaction as well as identifying the limiting reactant.

Moles to Grams Conversion

Converting between moles and grams is common in chemistry because it's often easier to measure a substance's mass than count individual molecules. To convert moles to grams, you need the molar mass of the substance, which is the mass of one mole of that substance.
The formula to use is:\[ \text{grams} = \text{moles} \times \text{molar mass} \]For methanol (\( \text{CH}_3\text{OH} \)), the molar mass is 32.04 g/mol. To find the theoretical yield in grams per minute, multiply the moles of methanol expected (from stoichiometry) by the molar mass. This conversion provides the theoretical yield in grams which can then be compared to the actual production to determine the percent yield.

• This comparison shows how efficiently the reaction is converting reactants to products.
• The percent yield calculation reveals real-world efficiency in contrast to theoretical potential.