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Question on Stoichiometry

Dhamnekar Winod

Active member
Nov 17, 2018
If any member knows the correct answer to the following question on stoichiometry, may reply to this question.
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Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
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, $10.0\,\text{dm}^3$.
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