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)?

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.

To solve this problem, we need to use the ideal gas law equation: PV = nRT. This equation relates the pressure (P), volume (V), number of moles (n), gas constant (R), and temperature (T). We can solve for any of these variables if we know the values of the other variables.

Given:
Mass of lithium metal (m) = 3.25g
Initial pressure (P1) = 0.991 atm
Initial temperature (T1) = 18°C = 18 + 273.15 = 291.15K
Final temperature (T2) = 41°C = 41 + 273.15 = 314.15K
Volume (V) = 235 mL = 235 cm³ = 235 x 10^-3 dm³

a) To find the pressure immediately after the completion of the reaction, we can assume that the volume and number of moles of gas remain constant in a closed system. Therefore, the equation can be rearranged as:

P1/T1 = P2/T2

P2 = (P1 * T2) / T1

Substituting the given values:

P2 = (0.991 atm * 314.15K) / 291.15K

P2 ≈ 1.067 atm

Therefore, the pressure in the flask immediately after the completion of the reaction is approximately 1.067 atm.

b) To find the pressure after the flask is cooled back down to room temperature, we can assume that the number of moles of gas remains constant, but the temperature changes. Rearranging the equation again:

P1/T1 = P2/T2

P2 = (P1 * T2) / T1

Substituting the given values:

P2 = (1.067 atm * 291.15K) / 314.15K

P2 ≈ 0.989 atm

Therefore, the pressure in the flask after it is allowed to cool back down to room temperature is approximately 0.989 atm.