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 \(25.0\) L/min. If \(5.30 \mathrm{~g}\) methanol is produced per minute, what is the percent yield of the reaction?

Short answer

The percent yield of methanol in the reaction is determined as follows: 1. Convert volumes of CO and H2 to moles: Moles of CO = \(\frac{25.0\,\text{L}}{22.4\,\text{L/mol}} = 1.12\,\text{mol}\) Moles of H2 = \(\frac{16.0\,\text{L}}{22.4\,\text{L/mol}} = 0.714\,\text{mol}\) 2. Determine the limiting reactant: Moles of methanol from CO = 1.12 mol Moles of methanol from H2 = \(0.714\,\text{mol}/2 = 0.357\,\text{mol}\) H2 is the limiting reactant as it can produce the least amount of methanol. 3. Calculate the theoretical yield of methanol: Theoretical yield of methanol (g) = \(0.357\,\text{mol} * 32.04\,\text{g/mol} = 11.43\,\text{g}\) 4. Calculate the percent yield: Percent yield = \(\frac{5.30\,\text{g}}{11.43\,\text{g}} * 100 = 46.37\%\) The percent yield of methanol in the reaction is 46.37%.

Step-by-step solution

Convert volumes of CO and H2 to moles

Since the gases are at standard temperature and pressure (STP), we can use the fact that \(1\,\text{mole}\) of a gas occupies a volume of \(22.4\,\text{L}\) at STP. For CO, we have: Moles of CO = volume of CO/Volume occupied by \(1\,\text{mole}\) For H2, we have: Moles of H2 = volume of H2/Volume occupied by \(1\,\text{mole}\)

Determine the limiting reactant

Based on the given number of moles of CO and H2, calculate the number of moles of methanol that can be produced by each reactant: Moles of methanol from CO = moles of CO Moles of methanol from H2 = moles of H2/2 The reactant that can produce the least amount of methanol is the limiting reactant.

Calculate the theoretical yield of methanol

The theoretical yield of methanol can be found by multiplying the moles of methanol produced by the limiting reactant by the molar mass of methanol (\(32.04\,\text{g/mol}\)). Theoretical yield of methanol (g) = moles of methanol produced * molar mass of methanol

Calculate the percent yield

The percent yield is calculated as the ratio of the actual yield (given as \(5.30\,\text{g}\)) to the theoretical yield, multiplied by 100. Percent yield = \(\frac{\text{actual yield}}{\text{theoretical yield}} * 100\) Now, plug in the values obtained in the previous steps to find the percent yield of methanol in the reaction.

Key concepts

Limiting Reactant

Understanding the concept of a limiting reactant is crucial in stoichiometry and for predicting the amounts of products formed in a chemical reaction. A limiting reactant is the substance that is used up first and limits the amount of product that can be formed. It's like the ingredient in a cookie recipe that runs out first—once it's gone, you can't make more cookies regardless of how much of the other ingredients you have left.

In the given exercise, we are provided with the rates of hydrogen and carbon monoxide being fed into a reactor. To identify the limiting reactant, we take into account the mole ratio between the reactants and the product, methanol. By comparing the amount of methanol that can be produced from each reactant's available moles, we can determine which reactant is the limiting one. This step is fundamental because it sets the stage for calculating the theoretical yield of the product.

Stoichiometry

Stoichiometry is the mathematical relationship between the quantities of reactants and products in a chemical reaction. It's based on the conservation of mass and the stoichiometric coefficients found in a balanced chemical equation. The key to solving stoichiometric problems is to convert quantities, such as masses or volumes, to moles, since moles directly reflect the proportional relationships outlined in the balanced equation.

In the exercise, stoichiometry helps us use the flow rates of reactants at STP to determine moles and then to calculate the quantities of product expected (theoretical yield). Improving exercise comprehension involves understanding how to convert between volume and moles for gases at STP, using the reaction's stoichiometry to compare reactants, and finding the theoretical yield based on the stoichiometry of the reactants and the limiting reactant.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be generated from a given amount of reactants, based on the stoichiometry of the reaction and assuming 100% efficiency. This is a hypothetical quantity calculated to predict the end point of a reaction if everything proceeds according to the balanced chemical equation with no losses.

In the context of the exercise, once we've established the limiting reactant, we can calculate the theoretical yield of methanol by using its molar mass. It's important to convey to students that the theoretical yield serves as a benchmark to gauge the efficiency of the actual reaction through the percent yield calculation. The comparison gives us insight into potential losses or inefficiencies in the reaction process.

Molar Volume at STP

Molar volume at standard temperature and pressure (STP) is the volume occupied by one mole of an ideal gas at 0°C (273.15 K) and 1 atmosphere of pressure. For all ideal gases, this volume is 22.4 liters (L). This makes calculations involving gases much simpler when they are at STP because we can directly relate volume of gas to the amount in moles.

In our example, this concept allows us to convert the given flow rates of hydrogen and carbon monoxide in liters per minute into moles. By using the molar volume at STP, we eliminate the need for more complex gas laws, streamlining the process of calculating the amount of reactants, which is essential for stoichiometry, determining the limiting reactant, and subsequently calculating the theoretical and percent yield of methanol.