Answer:
Percent yield =28.7%
Theoretical yield = 20.55 g
Explanation:
Given data:
Mass of Li₃N = ?
Mass of Li = 12.3 g
Mass of N₂ = 33.6 g
Actual yield = 5.89 g
Percent yield = ?
Solution:
Chemical equation:
6Li + N₂ → 2Li₃N
Number of moles of Li:
Number of moles = mass/molar mass
Number of moles = 12.3 g / 6.9 g/mol
Number of moles = 1.78 mol
Number of moles of N₂:
Number of moles = mass/molar mass
Number of moles = 33.6 g / 28 g/mol
Number of moles = 1.2 mol
Now we will compare the moles of both reactant with product.
N₂ : Li₃N
1 : 2
1.2 : 2/1×1.2 = 2.4 mol
Li : Li₃N
6 : 2
1.78 : 2/6×1.78 = 0.59 mol
Less number of moles of Li₃N are produced by Li thus it will be act as limiting reactant.
Mass of Li₃N: Theoretical yield
Mass = number of moles × molar mass
Mass = 0.59 mol × 34.83 g/mol
Mass = 20.55 g
Percent yield:
Percent yield = ( Actual yield / theoretical yield ) × 100
Percent yield = ( 5.89 g/ 20.55 g) × 100
Percent yield =28.7%
