What we will get if we mix 102ml H with 52ml O ?
I was reading "Gay Lussac’s Law of Gaseous Volumes", and was wondering about the equation.
I am a new chemistry teacher, so please bear with me.
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Assuming both those are in the form of gases, we know we would get water or H2O.
This is because of the law (can't remember whose) that one L of gas at the same temperature and pressure will always have the same amount of moles in it, no matter what the gas is.
So, we know that there is approximately double the amount of hydrogen compared to oxygen, so combining them together gives us H20 or water.
Then, depending on what the temperature is, the water might be in the form of liquid or gas.
Hope that helps...
I remember now:
It was Avogadro's Law. It states that equal volumes of gases under identical temperature and pressure conditions will contain equal numbers of particles (atoms, ion, molecules, electrons, etc.).
Going by the tenets of this law, one should expect to get 102 ml of gaseous product (here water). The liquified volume should be much lower, though. In addition to that the exact stoichiometry of the reaction H2 + 1/2 O2 ---> H2O requires 1/2 of 102 ml, i.e. 51 ml oxygen for this reaction. Hence there shold be 1 ml oxygen remining excessafter completion of reaction.
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