Question

If 34.2 grams of lithium react with excess water, how many liters of hydrogen gas can be produced at 299 Kelvin and 1.21 atmospheres? Show all of the work used to solve this problem.
2 Li (s) + 2 H2O (l) 2 LiOH (aq) + H2 (g)

Answer 1

Lithium becomes the limiting reagent here.

moles of lithium \[\frac{ 34.2g }{ 6.9 g mol ^{-1} } = 4.96 mol\]

Answer 2

From the equation, 2 moles of Lithium produces 1 mole of Hydrogen gas

2 mol Li --> 1 mol H2

4.96 mol Li --> (1/2) * 4.96 = 2.48 moles of Hydrogen gas

Answer 3

We then proceed to use the ideal gas equation, pV = nRT
p = 1.21 atm = 1.21 x 101325 = 122603 Pa
n = 2.48 mol
R = 8.31 J mol^-1 K^-1
T = 299

Using the equation we find that V = 0.0503 m^3 = 50.3 Litres

Answer 4

Why is the temperature 299 and why is the ideal constant (R) 8.31 instead of 0.0821?

Answer 5

the question asks how many litres of hydrogen can be produced at 299 Kelvin, so we need to set the value of T in the ideal equation to 299 K

I don't know what 0.0821 would represent, but the ideal gas constant is always equal to 8.31 J mol^-1 K^-1
The only time it would not be 8.31 would be if we introduced metric prefixes
So for example, R could equal to 8.31 J mol^-1 K^-1 but it could also be equal to 0.00831 kJ mol^-1 K^-1

Answer 6

Thanks!