Answer:
Volume of H2 produced = 26.5 L
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
[tex]Moles\ of \ Li \ present = \frac{Mass}{Atomic Mass} = \frac{21.3}{7} =3.04[/tex]
Therefore, moles of H2O produced will be:
[tex]=\frac{3.04\ moles\ Li*1\ mole\ H2O}{2\ moles\ Li} =1.52[/tex]
Step 2: Calculate the volume of H2 produced
Based on the ideal gas equation:
[tex]PV = nRT[/tex]
where P = pressure = 1.40 atm
T = Temperature = 297 K
n = number of moles = 1.52
R = gas constant = 0.0821 Latm/mol-K
[tex]V = \frac{nRT}{P} = \frac{1.52*0.0821*297}{1.40}=26.5\ L[/tex]
