Question

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

Answer 1

Hope this helps you.

Answer 2

Volume of H2 produced = 26.5 L

Step-by-step explanation:

Step 1: Calculate the moles of H2 produced

The given reaction is:

2Li + 2H2O → 2LiOH + H2

Here Li is the limiting reagent since H2O is in excess

Based on the reaction stoichiometry:

2 moles of Li produces 1 mole of water

Mass of Li = 21.3 g

Atomic mass of Li = 7 g/mol

`Moles\ of \ Li \ present = (Mass)/(Atomic Mass) = (21.3)/(7) =3.04`

Therefore, moles of H2O produced will be:

`=(3.04\ moles\ Li*1\ mole\ H2O)/(2\ moles\ Li) =1.52`

Step 2: Calculate the volume of H2 produced

Based on the ideal gas equation:

`PV = nRT`

where P = pressure = 1.40 atm

T = Temperature = 297 K

n = number of moles = 1.52

R = gas constant = 0.0821 Latm/mol-K

`V = (nRT)/(P) = (1.52*0.0821*297)/(1.40)=26.5\ L`