When 5.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of AlCl3 can be formed?
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Answer:
Final temperature is 302 K
Explanation:
You can now initial volume with ideal gas law, thus:
V =
Where:
n are moles: 2 moles
R is gas constant: 0,082
T is temperature: 300 K
P is pressure: 1 atm
V is volume, with these values: <em>49,2 L</em>
The work in the expansion of the gas, W, is: 1216 J - 34166 J = <em>-32950 J</em>
This work is:
W = P (Vf- Vi)
Where P is constant pressure, 1 atm
And Vf and Vi are final and initial volume in the expansion
-32950 J = -1 atm (Vf-49,2L) ×
Solving: <em>Vf = 49,52 L</em>
Thus, final temperature could be obtained from ideal gas law, again:
T =
Where:
n are moles: 2 moles
R is gas constant: 0,082
P is pressure: 1 atm
V is volume: 49,52 L
T is final temperature: <em>302 K</em>
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I hope it helps!