Combustion H2O(g) vs. H2O(l)

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Kayla Denton 1A
Posts: 106
Joined: Fri Sep 26, 2014 2:02 pm

Combustion H2O(g) vs. H2O(l)

Postby Kayla Denton 1A » Sat Jan 17, 2015 7:39 pm

In a combustion reaction, X + O2(g) --> CO2(g) + H2O(?). I was taught that H2O was gaseous (water vapour). However, in the solutions manual, H2O is written as H2O(l). This affects the answer to question 7.105, in which the change in moles of gas must be calculated for the reaction C6H6(l) + 15/2 O2(g) --> 6 CO2(g) + 3H2O(?). The solutions manual says that water is liquid, so the change in moles of gas is 6.00 - 7.50 = -1.50. I thought that H2O was gaseous, so I got 9.00 - 7.50 = +1.50.

Is there a rule dictating when H2O is gaseous and when it is liquid? Or is it always liquid?
Thanks!

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

Re: Combustion H2O(g) vs. H2O(l)

Postby Niharika Reddy 1D » Sun Jan 18, 2015 10:52 am

Unless stated otherwise (as in problem 115 where it says the products of the combustion reaction are gaseous), the products of the combustion of an organic compound are carbon dioxide gas and liquid water, and any nitrogen present is released as nitrogen gas (N2). If any other products are formed, the problem should specify that. I found this on the bottom of page 267 in the textbook for reference.


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