[mdawg467]

Givens
A student connects a flask containing 11.0 L of oxygen gas at 35^oC and 990 torr with another flask containing 15.0 L of hydrogen gas at 290 K and 1.40 atm.

Questions and My Attempt to Solve

Part A
What will the final pressure be if the stopcock connecting the two flasks is opened and the final temperature of the mixture is 310k?

From Pv=nrT,
n_h2= [(1.30atm)(11.0L)] / [(0.0821 Latm/Kmol)(308K)] = 0.566 mol H₂
n_O2= [(1.40atm)(15.0L)] / [(0.0821 Latm/Kmol)(290K)] = 0.882 mol O₂

n_total = n_H2 + n_O2 = 0.566 mol + 0.882 mol = 1.448 mol

From Pv=nrT,
P = [(1.448 mol)(0.0821 Latm/Kmol)(310K)] / [11.0 L + 15.0 L] = 1.42 atm

Part B
A spark ignites the mixture and a reaction occurs. Assuming that the reaction goes to completion to produce water gas and raise the final temperature to 55^oC. What is the final gas pressure?

2H₂(g) + O₂(g)  2H₂O (g)

Since every mol of oxygen requires 2 mols of hydrogen...
n_O2 = 0.882 mol
n_H2 = (2x0.566 mol) = 1.132 mol

n_total= n_O2 + (2)n_H2 = 2.014 mol

From Pv=nrT,
P = [(2.014 mol)(0.0821 Latm/Kmol)(328K)] / [26.0L] = 2.086 atm

Part C
If instead gaseous water, the pressure is great enough to produce liquid water and the final temperature is 45^oC. What is the final gas pressure?

Using n_total from part B..
From Pv=nrT,
P = [(2.014 mol)(0.0821 Latm/Kmol)(318K)] / [26.0 L] = 2.022 atm

Concerns
I'm not sure if I'm approaching Part B and Part C in the correct manner. Any help would be greatly appreciated.

Thank you.

[fledarmus]

Givens
A student connects a flask containing 11.0 L of oxygen gas at 35^oC and 990 torr with another flask containing 15.0 L of hydrogen gas at 290 K and 1.40 atm.

Questions and My Attempt to Solve


Part B
A spark ignites the mixture and a reaction occurs. Assuming that the reaction goes to completion to produce water gas and raise the final temperature to 55^oC. What is the final gas pressure?

2H₂(g) + O₂(g)  2H₂O (g)

Since every mol of oxygen requires 2 mols of hydrogen...
n_O2 = 0.882 mol
n_H2 = (2x0.566 mol) = 1.132 mol

You can't just arbitrarily double the amount of hydrogen in your mixture. You only have 0.566 moles of H₂ and 0.882 mol )₂ to work with. From your balanced equation, how much water can you form and what will be left over?