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Topic: Gas Problem  (Read 13678 times)

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Online Borek

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Re: Gas Problem
« Reply #15 on: June 26, 2007, 03:04:01 AM »
Note, that water on the right side of the equation should not be there - it is not a product, this water just happens to sit in the same place where your product (acetylene) is.

AWK: you are right it can be done this way. In fact that's how I calculated it at first. However, kemmy was all the time cofused and asking how to find water amount, so I decided to show him (her?) how to deal with the water problem.

BTW, have you noticed how proportionalities are often unknown to the posters?
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Offline billnotgatez

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Re: Gas Problem
« Reply #16 on: June 26, 2007, 03:21:13 AM »
Note, that water on the right side of the equation should not be there - it is not a product, this water just happens to sit in the same place where your product (acetylene) is.

CaC2 + 2H2O = C2H2(g) + Ca(OH)2(s)

This is technically correct - it just does not give me the feel for how wet participates in the meaning of the formula.


Online Borek

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Re: Gas Problem
« Reply #17 on: June 26, 2007, 07:40:14 AM »
Note, that water on the right side of the equation should not be there - it is not a product, this water just happens to sit in the same place where your product (acetylene) is.

CaC2 + 2H2O = C2H2(g) + Ca(OH)2(s)

This is technically correct - it just does not give me the feel for how wet participates in the meaning of the formula.

What if you did something like thermal decomposition of permanganate to oxygen? Just gas collecting over water. Same problem - you have to remember about water partial pressure if you want your calculations to be reasonably precise. And water is not present in the reaction equation on neither side.

Welcome to the real world ;)
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