A piece of lithium metal with a mass of 3.25g reacts with sufficient hydrochloric acid in a closed flask containing 235 mL of air originally at 0.991 atm and 18 C. The heat released in this reaction raises the temperature to 41.0 C inside the flask.

a)What is the pressure in the flask immediately after the completion of this reaction?

b)What is the pressure in the flask after it is allowed to cool back down to room temperature (18.0 C)?

1 answer

Do this:
2Li + 2HCl ==> H2 + 2LiCl
mols Li = 3.25/molar mass Li = 3.25/6.94 = 0.468 mols Li. That will produce 1/2 that of H2 so 0.468/2 = 0.234 mols H2 gas at 18 C. The volume of the flask is 235 mL.

Initially, before the reaction, calculate the mols of air and water vapor.
PV = nRT. Remember T is in Kelvin.
0.991*0.235 = n*0.08205*291
I obtained 0.00971 mols.

Immediately after the reaction, there will be 0.00971 mols air + water vapor + 0.234 mols H2. Add those together to obtain n. T is 273 + 41 = 314 K.
Use PV = nRT and calculate P for part a.
For part b, use PV = nRT with V, R, & n being the same but change T to 291 K and calculate P (for when the vessel has cooled).
Post your work if you get stuck.