Answer:
Number of moles of LiF produced by F₂ are less so it will limiting reactant.
Step-by-step explanation:
Given data:
Mass of lithium = 1.1 g
Mass of F₂ = 1.8 g
Limiting reactant = ?
Solution:
Chemical equation:
2Li + F₂ → 2LiF
Number of moles of Li:
Number of moles = mass/ molar mass
Number of moles = 1.1 g / 6.94 g/mol
Number of moles = 0.2 mol
Number of moles of F₂:
Number of moles = mass/ molar mass
Number of moles = 1.8 g / 38 g/mol
Number of moles = 0.05 mol
Now we will compare the moles of Li and F₂ with LiF.
Li : LiF
2 : 2
0.2 : 0.2
F₂ : LiF
1 : 2
0.05 : 2×0.05 = 0.1 mol
Number of moles of LiF produced by F₂ are less so it will limiting reactant.