/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Methanol, \(\mathrm{CH}_{3} \mat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified
135 g of hydrogen is required.

Step by step solution

01

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:\[\text{Mass of methanol} = \text{volume} \times \text{density} = 1000\, \text{mL} \times 0.791\, \text{g/mL} = 791\, \text{g}\]
02

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 \(\mathrm{H}_2\):\[\text{Molar mass of } \mathrm{H}_2 = 2.02\, \text{g/mol}\]\[\text{Mass of } \mathrm{H}_2 = 49.36\, \text{mol} \times 2.02 \frac{\text{g}}{\text{mol}} = 99.71\, \text{g}\]
03

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:\[\text{Actual mass of } \mathrm{H}_2 = \frac{99.71\, \text{g}}{0.74} \approx 134.75\, \text{g}\]
04

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

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 Hâ‚‚(g) → CH₃OH(â„“)
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 Hâ‚‚(g) → CH₃OH(â„“)
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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Sulfuric acid can be prepared starting with the sulfide ore, cuprite \(\left(\mathrm{Cu}_{2} \mathrm{S}\right) .\) If each \(\mathrm{S}\) atom in \(\mathrm{Cu}_{2} \mathrm{S}\) leads to one molecule of \(\mathrm{H}_{2} \mathrm{SO}_{4},\) what is the theoretical yield of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from \(3.00 \mathrm{kg}\) of \(\mathrm{Cu}_{2} \mathrm{S}\) ?

You have 0.954 g of an unknown acid, \(\mathrm{H}_{2} \mathrm{A},\) which reacts with NaOH according to the balanced equation $$\mathrm{H}_{2} \mathrm{A}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{A}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$$ If \(36.04 \mathrm{mL}\) of \(0.509 \mathrm{M} \mathrm{NaOH}\) is required to titrate the acid to the second equivalence point, what is the molar mass of the acid?

In some laboratory analyses, the preferred technique is to dissolve a sample in an excess of acid or base and then "back-titrate" the unreacted acid or base with a standard base or acid. To assess the purity of a sample of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) you dissolve a 0.475 -g sample of impure \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) in aqueous KOH. $$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{KOH}(\mathrm{aq}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{aq})+\mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$$ The \(\mathrm{NH}_{3}\) liberated in the reaction is distilled from the solution into a flask containing \(50.0 \mathrm{mL}\) of 0.100 M HCl. The ammonia reacts with the acid to produce \(\mathrm{NH}_{4} \mathrm{Cl},\) but not all of the HCl is used in this reaction. The amount of excess acid is determined by titrating the solution with standardized NaOH. This titration consumes \(11.1 \mathrm{mL}\) of 0.121 M NaOH. What is the weight percent of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) in the 0.475 -g sample?

What volume of \(0.750 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\), in milliliters, is required to react completely with 1.00 L of \(2.25 \mathrm{M} \mathrm{NaCl}\) solution? The balanced equation is \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaCl}(\mathrm{aq}) \rightarrow \mathrm{PbCl}_{2}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})\)

Oyster beds in the oceans require chloride ions for growth. The minimum concentration is \(8 \mathrm{mg} / \mathrm{L}\) (8 parts per million). To analyze for the amount of chloride ion in a 50.0 -mL sample of water, you add a few drops of aqueous potassium chromate and then titrate the sample with \(25.60 \mathrm{mL}\) of 0.001036 M silver nitrate. The silver nitrate reacts with chloride ion, and, when the ion is completely removed, the silver nitrate reacts with potassium chromate to give a red precipitate. (a) Write a balanced net ionic equation for the reaction of silver nitrate with chloride ions. (b) Write a complete balanced equation and a net ionic equation for the reaction of silver nitrate with potassium chromate, indicating whether each compound is water-soluble or not. (c) What is the concentration of chloride ions in the sample? Is it sufficient to promote oyster growth?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.