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Nov 18, 2023 at 9:53 comment converted from answer user317384 @Karl I really don't understand the criticism of the question. Surely students should learn to calculate yield via stoichiometry? It's not a fictional process: "The industrial production of synthetic methanol abroad has been known for some time and in May, 1927, an industrial production of 4500 gallons of methanol daily was announced by one American company. In this instance, carbon dioxide and hydrogen, hitherto waste gases from the production of butanol and acetone from corn, were utilized." Methanol from Hydrogen and Carbon Monoxide, 1928
Jan 13, 2023 at 9:11 comment added Poutnik I have evaluated the attempt 1 some time ago and it does seem to me to be an error in the task formulation or answer list preparation. If we considered hydrogen as a real gas at this high pressure, the resulting volume would be even higher.
Jan 13, 2023 at 8:33 history edited andselisk CC BY-SA 4.0
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Jan 13, 2023 at 5:06 history edited Jane902 CC BY-SA 4.0
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Jan 10, 2023 at 4:15 comment added Jane902 @Maurice I thought it might be an error in the mark scheme but 9.1 L is not even one of the MCQ options so it couldn’t have been a typo. Like I mentioned earlier it’s from a past preliminary Olympiad test. Usually the questions are fully vetted. I really want confirm that it’s an error in the question and my answer is right before getting back to the student who asked me this.
Jan 10, 2023 at 4:02 comment added Jane902 @CurtF. What do you mean? I used density=mass/volume to find the mass of methanol and I used that mass in the yield calculation
Jan 10, 2023 at 4:01 comment added Jane902 @Karl This question is from a past paper of a Chemistry Olympiad test. You’re right I don’t think they considered how practical this question is.
Jan 9, 2023 at 19:45 comment added Maurice Your first calculation was right. The volume of $\ce{H2}$ must be nearly $9$ L, and not nearly $5$ L. $5$ L is not far from the expected volume of $\ce{CO}$.
Jan 9, 2023 at 16:40 comment added Curt F. I think the yield has units of mols per mol (i.e. actual moles per theoretical moles is 0.95), not grams per gram. I don't think it makes a difference numerically but it'd be cleaner to convert to moles first, then adjust for yield.
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Jan 9, 2023 at 1:12 comment added Jane902 @Poutnik Yeah my attempt 2 is definitely wrong, couldn't think of anything else though. I don't really get why attempt 1 is wrong. What am I missing?
Jan 9, 2023 at 1:09 history edited Jane902 CC BY-SA 4.0
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Jan 9, 2023 at 1:09 comment added Jane902 @matt_black Yeah sorry it should be 5 L
Jan 8, 2023 at 20:01 comment added Poutnik @Natalie Methanol at 20 deg C is liquid, not gas, and you cannot use for it the ideal gas state equation. (attempt 2 )
Jan 8, 2023 at 19:57 comment added Poutnik @matt_black Take the bold text. approximately how many liters of hydrogen gas at 350 °C and 300 atm
Jan 8, 2023 at 19:53 comment added matt_black @Poutnik The wording for attempt 2 says "Assuming the pressure of methanol is 1 atm" so maybe it is careless wording, but sounds wrong.
Jan 8, 2023 at 19:50 comment added Poutnik @matt_black It is volume of pressurized (300 atm) hydrogen, not methanol.
Jan 8, 2023 at 18:16 comment added matt_black If the correct answer is d) the volume of methanol is 5 L not 5mL. And that is the volume of the liquid so no calculations using the gas equation are relevant to it.
Jan 8, 2023 at 16:51 comment added Poutnik It is always good to summarize and/or complement the verbal description by algebra. Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers. It helps in focusing on principles, mistakes are easier to spot, orientation is improved, Q/A is reusable and has bigger permanent value. You may find useful formatting mathematical/chemical expressions/formulas.
Jan 8, 2023 at 16:13 history edited Jane902 CC BY-SA 4.0
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