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2 mole of each of O2 and H2 are in seperate containers each of volume Vo and at 150C and 1 atmosphere. the 2 are made to react in a third container to form water vapour until H2 is exhausted. when temperature of the mixture in the third container was restored top 150C its pressure becomes 1 atmosphere the volume of third container must be 1. Vo 2. 5Vo/2 3. 3Vo/2 4. 2Vo
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look at the equation \(O_2+2H_2\rightarrow 2H_2O\) if you had \(V_{\circ}\) of \(H_2\), then the volume of \(H_2O\) you can make is only \(V_{\circ}\) (note their coefficients). Are you also supposed to take the volume of \(O_2\) into account? if so you need to add \(\dfrac{1}{2}V_{\circ}\) to the volume of \(H_2O\) vapour.
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