A rigid stainless steel chamber contains 210 Torr of methane, CH4, and excess oxygen, O2, at...

A rigid stainless steel chamber contains 210 Torr of methane, CH4, and excess oxygen, O2, at 200.0

Expert Solution

CH4 + 2O2 ---> CO2 + 2H2O

CH4 + 2O2 = CO2 + 2H2O

So 1 volume of CH4 + 2 volumes of O2 yields 1 volume of CO2 + 2 volumes of H2O

210 torr of CH4 pressure means at least 420 torr of O2 pressure. Since it is excess, let us call the O2 pressure 420 + x.

Let us consider the CH4 to CO2 molar ratio of 1 to 1. Since this happens at constant volume and constant temperature, the pressures are directly related to the moles used and produced. Thus, we conclude 210 torr of CO2 pressure is produced.

Since we are at 200 C, the water produced is gaseous. Let us consider the CH4 to H2O molar ratio of 1 to 2. This informs us that 420 torr of water vapor pressure is created.

Since O2 is in excess, we know that x remains after 420 torr of O2 is used up. I know 420 torr of O2 is used because of the CH4 to O2 molar ratio in the balanced equation.

I think the water will remain as vapor given that the normal boiling point of water is 100 C at 760 torr.

Conclusion: no pressure change in the chamber at reaction's end.

queen_honey_blossom answered 1 year ago

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