a mixture of 60cm3 of H2 and 40cm3 CO is passed over excess copper(II) Oxide
calculate volume and composition of resulting gas anyhelp would be much appreciated
1
composition of gas
Started by xClairex, Mar 22 2007 09:57 PM
2 replies to this topic
#1
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Posted 22 March 2007 - 09:57 PM
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Posted 22 March 2007 - 11:24 PM
QUOTE(xClairex @ Mar 22 2007, 09:57 PM) 
a mixture of 60cm3 of H2 and 40cm3 CO is passed over excess copper(II) Oxide
calculate volume and composition of resulting gas anyhelp would be much appreciated
calculate volume and composition of resulting gas anyhelp would be much appreciated
Both H2 and CO are good reducing agents, meaning that they will be able to remove the oxygen from copper(II) oxide (and convert copper ions to copper atoms).
In the process H2 will be converted into H2O on a one mole to one mole basis.
Assuming we are talking about normal temperature, the H2O will eventually become liquid so it will contribute 0 cm3 to the final volume.
The CO will turn into CO2, again on a one mole to one mole basis.
The 40 cm3 of CO will convert into 40 cm3 of CO2.
As far as I can see from the information you have supplied, that will be the only gas present.
Hope this helps.
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Posted 23 March 2007 - 11:51 PM
thank you i understand more now 
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