Problem 24 - Chemistry and Chemical Reactivity

Question

Methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) can be prepared from carbon monoxide and hydrogen. $$\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\ell)$$ What mass of hydrogen is required to produce 1.0 L of \(\mathrm{CH}_{3} \mathrm{OH}(d=0.791 \mathrm{g} / \mathrm{mL})\) if this reaction has a \(74 \%\) yield under certain conditions?

Step-by-Step Solution

Answer

135 g of hydrogen is required.

Step 1: Calculate the Mass of Methanol Needed

First, calculate the mass of methanol (\[\mathrm{CH}_3\mathrm{OH}\]) that is needed. The volume of methanol given is 1.0 L. Since the density is 0.791 g/mL, we convert the volume from liters to milliliters (1 L = 1000 mL). Thus, the mass of methanol is calculated as follows:\[\text{Mass of methanol} = \text{volume} \times \text{density} = 1000\, \text{mL} \times 0.791\, \text{g/mL} = 791\, \text{g}\]

Step 2: Calculate the Theoretical Mass of Hydrogen Needed

Calculate the theoretical mass of hydrogen gas (\[\mathrm{H}_2\]) needed assuming a 100% yield of methane. First, we write the balanced chemical equation:\[\mathrm{CO}(\mathrm{g}) + 2 \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{CH}_3\mathrm{OH}(\ell)\]This shows that 2 moles of \(\mathrm{H}_2\) produce 1 mole of \(\mathrm{CH}_3\mathrm{OH}\). Next, find the moles of methanol needed:\[\text{Molar mass of } \mathrm{CH}_3\mathrm{OH} = 12.01 + 3\times1.01 + 16.00 + 1.01 = 32.05\, \text{g/mol}\]\[\text{Moles of } \mathrm{CH}_3\mathrm{OH} = \frac{791\, \text{g}}{32.05\, \text{g/mol}} \approx 24.68\, \text{mol}\]Thus, moles of \(\mathrm{H}_2\) required:\[\text{Moles of } \mathrm{H}_2 = 2 \times 24.68 = 49.36\, \text{mol}\]Finally, calculate the mass of \(\mathrm{H}_2\):\[\text{Molar mass of } \mathrm{H}_2 = 2.02\, \text{g/mol}\]\[\text{Mass of } \mathrm{H}_2 = 49.36\, \text{mol} \times 2.02 \frac{\text{g}}{\text{mol}} = 99.71\, \text{g}\]

Step 3: Adjust for the Reaction Yield

Since the reaction yield is only 74%, we need to adjust the mass of \(\mathrm{H}_2\) to reflect this yield. Calculate the actual mass of hydrogen required by dividing the theoretical mass by the yield:\[\text{Actual mass of } \mathrm{H}_2 = \frac{99.71\, \text{g}}{0.74} \approx 134.75\, \text{g}\]

Step 4: Final Answer

To produce 1.0 L of methanol with a density of 0.791 g/mL and a yield of 74%, the mass of hydrogen required is approximately 135 g.

Key Concepts

Understanding Stoichiometry

Stoichiometry is a key concept in chemistry that helps us understand the quantitative aspects of chemical reactions. When dealing with reactions, stoichiometry provides a way to determine the amounts of reactants needed to create a specific amount of product. This involves using a balanced chemical equation, which specifies the exact ratio between reactants and products.

For example, in the methanol production reaction

• CO(g) + 2 H₂(g) → CH₃OH(ℓ)

This tells us that 1 mole of carbon monoxide reacts with 2 moles of hydrogen to produce 1 mole of methanol.

By understanding stoichiometry, we can calculate how much of each reactant is needed or how much product can be formed from given reactants. It requires knowledge of molecular weights and the ability to convert between grams and moles using these ratios.

In our case, we start by calculating the moles of methanol formed using its molar mass. Then, using stoichiometry, we determine how many moles of hydrogen gas are required based on the balanced equation.

Methanol Production Process

Methanol, also known as wood alcohol, is produced through a chemical reaction involving carbon monoxide and hydrogen gas. This process is an example of synthesis, where simpler substances combine to form a more complex compound. The reaction is as follows:

• CO(g) + 2 H₂(g) → CH₃OH(ℓ)

Methanol production is crucial in the chemical industry as it serves as a base material for many other chemical compounds.

The need for precise calculations in methanol production is critical because industrial setups often require maximizing efficiency and minimizing waste. In our example, the yield of the reaction isn't always perfect due to various factors like reaction conditions or impurities. That's why calculating the yield and adjusting reactant amounts based on this yield is essential for efficient production.

Calculating the theoretical yield allows us to determine the maximum amount of product possible under ideal conditions. However, the actual yield, like our 74%, tells us what was practically achieved, requiring more reactants to compensate for less-than-ideal conversion.

Importance of Density Calculation

Density is a measure of mass per unit volume and plays a vital role in understanding the properties of substances. In our exercise, the density of methanol is given as 0.791 g/mL, crucial for converting between volume and mass—a necessary step in our calculations.

To find out how much methanol needs to be produced, we start from the volume given (1.0 L) and convert it to mass using the density value. This conversion is straightforward: multiplying the volume by the density provides the mass of methanol in grams.

The formulas used for this purpose are:

• Mass = Volume × Density

Understanding density isn't just helpful in lab calculations; it's also important in real-world applications, like knowing how to store chemicals safely based on their densities or understanding flow properties. It helps bridge the gap between how much of a substance you have in your laboratory beaker versus how much it weighs.