Problem 70 - Chemistry The Molecular Science

Question

Methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) is a clean-burning, easily handled fuel. It can be made by the direct reaction of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\). $$\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(\ell)$$ (a) For a mixture of \(12.0 \mathrm{~g} \mathrm{H}_{2}\) and \(74.5 \mathrm{~g} \mathrm{CO},\) determine the limiting reactant. (b) Calculate the mass (g) of the excess reactant left after reaction is complete. (c) Calculate the theoretical mass of methanol that can be obtained.

Step-by-Step Solution

Answer

The limiting reactant is CO, 1.25 g of H2 is left, and 85.2 g of methanol is produced.

Step 1: Convert to moles of CO

First, determine the moles of CO present in the mixture. Use the molecular weight of CO, which is 28.01 g/mol. Calculate the moles by dividing the given mass by the molecular weight:\[\text{moles of CO} = \frac{74.5 \, \text{g}}{28.01 \, \text{g/mol}} \approx 2.66 \, \text{moles}\]

Step 2: Convert to moles of H2

Next, calculate the moles of H2 using its molecular weight, 2.02 g/mol:\[\text{moles of } \mathrm{H}_{2} = \frac{12.0 \, \text{g}}{2.02 \, \text{g/mol}} \approx 5.94 \, \text{moles}\]

Step 3: Determine limiting reactant

Use the stoichiometry of the reaction. According to the equation, 1 mole of CO reacts with 2 moles of H2. Compare the available moles:Given 2.66 moles of CO, you would need:\[2.66 \times 2 = 5.32 \, \text{moles of } \mathrm{H}_{2}\]Since we have 5.94 moles of H2, and require only 5.32 moles of H2, CO is the limiting reactant as all available CO moles will be used first.

Step 4: Calculate excess H2 remaining

Calculate moles of H2 needed:\[\text{moles of } \mathrm{H}_{2} \, \text{used} = 2.66 \times 2 = 5.32 \, \text{moles}\]Subtract from total moles of H2:\[\text{excess moles of } \mathrm{H}_{2} = 5.94 - 5.32 = 0.62 \, \text{moles}\]Convert moles to grams:\[\text{mass of excess } \mathrm{H}_{2} = 0.62 \, \text{moles} \times 2.02 \, \text{g/mol} \approx 1.25 \, \text{g}\]

Step 5: Calculate theoretical mass of methanol

The reaction produces 1 mole of methanol per mole of CO used:Moles of methanol = moles of limiting reactant, which is CO:\[2.66 \, \text{moles of } \mathrm{CH}_{3} \mathrm{OH}\]Convert to mass using molar mass of methanol, 32.04 g/mol:\[\text{mass of } \mathrm{CH}_{3} \mathrm{OH} = 2.66 \times 32.04 \, \text{g/mol} \approx 85.2 \, \text{g}\]

Key Concepts

Limiting Reactant

When conducting chemical reactions, the concept of the limiting reactant is crucial for understanding which substance will determine the amount of product that can be formed. In any reaction, the limiting reactant is the reactant that is completely consumed first, thereby stopping the reaction from continuing.
In this exercise, we identify the limiting reactant by comparing the mole ratio of the reactants, based on the balanced chemical equation:

1. According to the stoichiometry of the equation, 1 mole of carbon monoxide (CO) reacts with 2 moles of hydrogen (H\(_2\)) to produce methanol.
2. By calculating, we find that 74.5 g of CO converts to 2.66 moles, while 12.0 g of H\(_2\) converts to about 5.94 moles.
3. Since 2.66 moles of CO would require exactly 5.32 moles of H\(_2\) (calculated as 2.66 moles of CO multiplied by the stoichiometric factor of 2), and 5.94 moles are available, CO is consumed entirely, making it the limiting reactant.

Understanding which is the limiting reactant is essential because it determines how much of the final product can be theoretically produced.

Excess Reactant

The excess reactant is the substance that is not completely used up when the reaction comes to an end. Identifying the excess reactant helps in calculating any leftover materials post-reaction, which can be useful for recycling or efficiency improvement.

• In the given exercise, after establishing CO as the limiting reactant, we know that hydrogen will remain unreacted once all the CO is consumed.


• Starting with 5.94 moles of H\(_2\) and using 5.32 moles to react with the available CO, we determine that 0.62 moles of H\(_2\) are left over.


• To find out how much excess reactant remains by mass, these 0.62 moles of H\(_2\) are converted to grams: 0.62 moles × 2.02 g/mol, resulting in approximately 1.25 g of H\(_2\).

This leftover amount of hydrogen indicates an efficient use of reactants, but also an opportunity for optimization during multiple reaction cycles.

Theoretical Yield

Theoretical yield is the maximum amount of product that can be synthesized from a given amount of reactant, assuming complete utilization of the limiting reactant. Calculating the theoretical yield provides insight into the perfect efficiency scenario of the reaction.

• In the case of this reaction, since carbon monoxide (CO) is the limiting reactant, the moles of methanol (CH\(_3\)OH) produced will equal the moles of CO used. For 2.66 moles of CO, a corresponding 2.66 moles of methanol should form.


• To convert moles of methanol to mass, use its molar mass of 32.04 g/mol. Thus, the theoretical mass of methanol is 2.66 moles × 32.04 g/mol, which results in approximately 85.2 g of methanol.

This theoretical yield represents the upper limit of methanol formation if the reaction proceeds perfectly, without losses or side reactions, providing a benchmark for evaluating actual experimental results.