Chemistry Question -- Opening a valve between two containers....

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SUMMARY

The discussion focuses on calculating the moles of hydrogen (H2) and oxygen (O2) gases in two separate containers before a valve is opened. Container A contains 8.00 L of H2 at 3.00 atm and 400 K, resulting in 0.731 moles of H2 using the ideal gas law (PV = nRT). Container B holds 6.00 L of O2 at 4.00 atm and 400 K, yielding 0.732 moles of O2. The reaction between the gases to form water (H2O) is not relevant to the calculations prior to opening the valve.

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Homework Statement


Two closes containers are connected by a valve, which initially separates the containers from each other. C
Container A has 8.00 L of H2 at 3.00 atm and temperature 400 k.
Container B has 6.00 L of O2 at 4.00 atm and temperature 400 k

The valve is opened, the gases mix, and they then react to form H2O. (temperature for both stay the same)
Find the moles of each gases before the valve is opened.

Homework Equations


Chemical formula: 2 H2 (g) + O2 (g) === 2 H2O (g)

The Attempt at a Solution


for H2: PV/ RT , (3.00 atm) (8.00 L) / (0.0821 (L atm/ mol k)) (400 k) = 0.731 mol H2

I also got the same moles for O2, but i feel like i did something wrong. Can anyone help me?
 
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You did OK, the part about the reaction is irrelevant to the system before the valve was open.
 

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