If 21.3 grams of lithium react with excess water, how many liters of hydrogen gas can be produced at 297 Kelvin and 1.40 atmospheres? Show all of the work used to solve this problem. 2Li (s) + 2H2O (l) 2LiOH (aq) + H2 (g)
2Li(s) + 2H2O(l) --> 2LiOH(aq) + H2(g) 21.3g Li /7g/mole = 3.04moles Li 3.04moles Li x (1H2 / 2Li) = 1.52moles H2 pv = nrt v = nrt/p v = 1.52moles x 0.0821L-atm/mole-K x 297K / 1.4atm v = 26.47L
