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The discussion centers on a stoichiometry question regarding gas volumes during a chemical reaction. It establishes that the volume of a gas at standard pressure and temperature (STP) is determined by the number of molecules, leading to the conclusion that the volume of sulfur trioxide ($\ce{SO3(g)}$) remains at $10.0\,\text{dm}^3$ after the reaction. The initial volumes of sulfur dioxide ($\ce{SO2(g)}$) and oxygen ($\ce{O2(g)}$) are $10.0\,\text{dm}^3$ and $6.0\,\text{dm}^3$, respectively, with only $5.0\,\text{dm}^3$ of oxygen being consumed. This confirms the principle of gas volume conservation in stoichiometric reactions.

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WMDhamnekar
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Hi,
If any member knows the correct answer to the following question on stoichiometry, may reply to this question.
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Hey Dhamnekar,

The volume of a gas only depends on the number of molecules given the same pressure and temperature (give or take a little).
It's why we have a molar volume at standard pressure and temperature ($V_m$), which is the volume of 1 mole of gas irrespective of which gas it is.

The volume of $\ce{SO2(g)}$ before the reaction is $10.0\,\text{dm}^3$.
The volume of $\ce{O2(g)}$ is $6.0\,\text{dm}^3$ of which only $5.0\,\text{dm}^3$ will be used (half of $10.0\,\text{dm}^3$).
After the reaction we have the same number of molecules of $\ce{SO3(g)}$, so its volume will be the same. That is, $10.0\,\text{dm}^3$.
 
Last edited:

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