Gay Lussac and Avogadro's Laws (1 Viewer)
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Jaydels
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Gay Lussac's Law
"the ratio of the volumes of gases involved in a reaction if measured at the same temperature and pressure, are expressed by small, whole numbers"
Avagadro's Law
"equal volumes of gases measured temperature and pressure, contain equal numbers of molecules"
eg. in the equation 2H2 + O2 --> 2H20
in 2L of H2 there must be the same number of molecules as in 2L of O2, therefore, in 2L of H2 there must be twice the number of molecules there are in 1L of O2, providing they are the same temperature and pressure.
In your example there are 13 moles of O2 and 2 moles of C4H10, so if you have 40ml of butane, you will need 13/2 x 40ml of oxygen to completely burn it (because there are 13/2 times more molecules in oxygen), which will produce 8/2 x 40ml of carbon dioxide.
I hope that helps a bit...
"the ratio of the volumes of gases involved in a reaction if measured at the same temperature and pressure, are expressed by small, whole numbers"
Avagadro's Law
"equal volumes of gases measured temperature and pressure, contain equal numbers of molecules"
eg. in the equation 2H2 + O2 --> 2H20
in 2L of H2 there must be the same number of molecules as in 2L of O2, therefore, in 2L of H2 there must be twice the number of molecules there are in 1L of O2, providing they are the same temperature and pressure.
In your example there are 13 moles of O2 and 2 moles of C4H10, so if you have 40ml of butane, you will need 13/2 x 40ml of oxygen to completely burn it (because there are 13/2 times more molecules in oxygen), which will produce 8/2 x 40ml of carbon dioxide.
I hope that helps a bit...
Emma-Jayde
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The volume ratios have to stay the same. For
C4H10 + 13/2 O2 ---> 4 CO2 + 5 H2O
(I often find it easier to balance diatomic substances in halves, It means you don't have to halve everything later)
You have 1 of C4H10, so therefore you need 13/2 volumes of O2 for every volume of C4H10. The same goes for the CO2. You need 4 volumes of CO2 for every volume of C4H10.
You burned 40 units of butane, so therefore you need 13/2 times 40 units of oxygen, and 4 times 40 units of carbon dioxide, giving you 260 units of oxygen and 160 units of carbon dioxide.
This will give you
40 C4H10 + 260 O2 ---> 160 CO2 + 200 H2O
See how the ratio of volumes stays the same? If you divide through be 40, you'll get the smae volumes as in the 1st equation
Hope this helps. If you didn't understand, say exactly what it is you didn't get
C4H10 + 13/2 O2 ---> 4 CO2 + 5 H2O
(I often find it easier to balance diatomic substances in halves, It means you don't have to halve everything later)
You have 1 of C4H10, so therefore you need 13/2 volumes of O2 for every volume of C4H10. The same goes for the CO2. You need 4 volumes of CO2 for every volume of C4H10.
You burned 40 units of butane, so therefore you need 13/2 times 40 units of oxygen, and 4 times 40 units of carbon dioxide, giving you 260 units of oxygen and 160 units of carbon dioxide.
This will give you
40 C4H10 + 260 O2 ---> 160 CO2 + 200 H2O
See how the ratio of volumes stays the same? If you divide through be 40, you'll get the smae volumes as in the 1st equation
Hope this helps. If you didn't understand, say exactly what it is you didn't get
Jaydels
Member
no probs 
I wasn't sure whether mine made sense...but anyway
I wasn't sure whether mine made sense...but anyway
Emma-Jayde
Muahahahaha
- Joined
- Mar 5, 2005
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- 2005
Lol, I wasn't too sure if I made sense either...Well it made sense to ME...Not that that means much!