Question

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When heated, lithium reacts with nitrogen to form lithium nitride:

Li(s) + N2(g) ----------> Li3N(s)

What is the theoretical yield of Li3N in grams when 12.3 g of Li are heated with 33.6 g of N2?

If the actual yield of Li3N is 5.89 g, what is the percent yield of the reaction?​

Answer 1

Percent yield =28.7%

Theoretical yield = 20.55 g

Step-by-step 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%