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Question

A mixture in which the mole ratio of H2 and O2 is :1 is used to prepare water by the reaction,
2H2(g)+O2(g)2H2O(g)
The total pressure in the container is 0.8 atm at 20oC before the reaction. Determine the final pressure at 120oC after reaction assuming 80% yield of water.

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Solution

PH2=23×0.8=0.533atm

PO2=13×0.8=0.244atm

2H2+O22H2O
i=0 0.533 0.266 0

After the reaction 0.533×201000.266×201000.533×80100

Total pressure =0.1066+0.0533+0.4264=0.5863 atm

Using Gay-Lussac's law, P1T1=P2T2
0.5863293=P2393

P2=0.7864atm

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