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:
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
\(Moles\ of \ Li \ present = \frac{Mass}{Atomic Mass} = \frac{21.3}{7} =3.04\)
Therefore, moles of H2O produced will be:
\(=\frac{3.04\ moles\ Li*1\ mole\ H2O}{2\ moles\ Li} =1.52\)
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
\(V = \frac{nRT}{P} = \frac{1.52*0.0821*297}{1.40}=26.5\ L\)