If 24.6 grams of lithium react with excess water, how many liters of hydrogen gas can be produced at 301 Kelvin and 1.01 atmospheres? Show all of the work used to solve this problem.

2 Li (s) + 2 H2O (l) 2 LiOH (aq) + H2 (g)

Ok here is how we are going to do it:

First we need to know that lithium's molecular weight is 6.941 g/mol.

24.6 g of Li x (1 mol of Li / 6.941 g of Li) = 3.544 mol Li

3.544 moles Li x (1 moles H2/2 moles Li) = 1.772 mol H2

n = 1.772 moles P = 1.01 atm T = 301 K

R = 0.08205746 (L atm) / (mol K).

V = nRT/P = (1.772 moles * 0.08205746 (L atm) / (mol K) x 301 K) / 1.01 atm = 43.33 Liters of H2