Question

What is the theoretical yield of Li3N in grams when 12.3 g of Li is heated with 33.6g of N2?if the actual yield of Li3N is 5.89g,what is the percent yield of thereaction?

Answer 1

In order to determine the theoretical yield, we must first find the limiting reactant (reagent), which will determine the greatest possible amount of product that can be produced.

Molar Masses
$\text{Li}:$$\text{6.941 g/mol}$
${\text{N}}_2:$$(2\times 14.007\text{g/mol})=\text{28.014 g/mol}$
${\text{Li}}_3\text{N}:$$(3\times 6.941\text{g/mol Li})+(1\times 14.007\text{g/mol N})={\text{34.83 g/mol Li}}_3\text{N}$

Limiting Reactant
Divide the mass of each reactant by its molar mass, then multiply times the mole ratio from the balanced equation with the product on top and the reactant on bottom, then multiply times the molar mass of ${\text{Li}}_3\text{N}$.

Lithium
$12.5\text{g Li}\times \frac{1\text{mol Li}}{6.941\text{g Li}}\times \frac{2{\text{mol Li}}_3\text{N}}{6\text{mol Li}}\times \frac{34.83{\text{g Li}}_3\text{N}}{1{\text{mol Li}}_3\text{N}}={\text{20.9 g Li}}_3\text{N}$

Nitrogen Gas
$34.1{\text{g N}}_2\times \frac{1{\text{mol N}}_2}{28.014{\text{g N}}_2}\times \frac{2{\text{mol Li}}_3\text{N}}{1{\text{mol N}}_2}\times \frac{34.83{\text{g Li}}_3\text{N}}{1{\text{mol Li}}_3\text{N}}={\text{84.8 g Li}}_3\text{N}$

Lithium produces less lithium nitride than nitrogen gas. Therefore, the limiting reactant is lithium, and the theoretical yield of lithium nitride is $\text{20.9 g}$.

Answer 2

6 Li + N2 => 2 Li3N

Theoretical moles of Li : N2 = 6 : 1

Moles of Li = mass/molar mass of Li

= 12.3/6.941 = 1.772 mol

Moles of N2 = mass/molar mass of N2

= 33.6/28.01 = 1.200 mol

Experimental moles of Li : N2 = 1.772 : 1.200

= 1.477 : 1 = 6 : 4.063

Since N2 is in excess, Li is the limiting reactant

Moles of Li3N = 2/6 x moles of Li

= 2/6 x 1.772 = 0.5907 mol

Theoretical yield of Li3N = moles x molar mass of Li3N

= 0.5907 x 34.83

= 20.6 g

Percent yield = actual yield/theoretical yield x 100%

= 5.89/20.6 x 100%

= 28.6%