Let's say I have a mixture of $\ce{CO2}$ and $\ce{H2O}$ gases. I've read that $\ce{CO2}$ is heavier (molar mass is $\pu{44.01 g/mol}$) than $\ce{H2O}$ (molar mass is $\pu{18.02 g/mol}$). So, if the mixture is supplied to a cylinder from the bottom, will only the water vapour reach the top? If so, is this a good way to separate the $\ce{CO2}$ and water vapour? Or is there a more efficient way to separate them?
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$\begingroup$ Convenient reference for text/formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // For more: Math SE MathJax tutorial. // Not to be applied in CH SE titles. $\endgroup$– PoutnikNov 29, 2022 at 14:33
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1$\begingroup$ What you propose does not work at all. Never. $\endgroup$– KarlNov 29, 2022 at 19:38
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$\begingroup$ @Karl Then, could you please write an answer for a better solution to this problem? $\endgroup$– Ishaan ManishNov 30, 2022 at 7:14
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$\begingroup$ It is hard to see any source containing a mixture of CO2 and water as gasses. Most likely under normal conditions you would have water liquid under an atmosphere of carbon dioxide (with the water saturated with CO2). Or a stream of hot gases containing both. What circumstances are you considering? $\endgroup$– matt_blackNov 30, 2022 at 14:34
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2 Answers
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Separation of $\ce{H2O}$ from $\ce{CO2}$ has been done by all analytical chemists in the $19$th century, when they had to determine the formula of a new organic compound X. The compound X was mixed with some $\ce{CuO}$ powder, introduced in a horizontal glass tube and heated by a burner in an oxygen current. X was burned int $\ce{CO2}$ and $\ce{H2O}$. Then the oxygen gas current containing some $\ce{H2O}$ and $\ce{CO2}$ had to cross another glass tube filled with pieces of anhydrous calcium chloride $\ce{CaCl2}$ which fixed the water vapor, producing $\ce{CaCl2·6H2O}$. The $\ce{CaCl2}$ tube was weighed before and after the operation. The increase in weight gives the total amount of $\ce{H2O}$ in the mixture of gases.
Then the gas was sent to bubble through a concentrated $\ce{NaOH}$ solution, where the following reaction happened $$\ce{CO2 + 2 NaOH -> Na2CO3 + H2O}$$ The $\ce{NaOH}$ solution was weighed before and after the operation. The increase in weight gives the total amount of $\ce{CO2}$ in the gas.
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It depends on what equipment you have available. A mix of compression and cooling will work much better than relying on their molecular weights.
Water boils at 100 degC, while CO2 sublimates at -78 degC. Simple cooling will ensure that water condenses long before the CO2, while compression will drive the formation formation of water-water bonds. The Clapeyron equation should let you calculate the needed values.
