Question

Lithium Nitrogen reacts to form Li3N. A 4.0g sample of lithium is reacted with 2.5g of nitrogen. What weight of what substance remains unreacted?

Answer 1

The limiting reactant is the reactant that is completely used up during the reaction. Its opposite, the excess reactant, is the reactant that is not completely reacted.


We are given 4.0g of Li and 2.5g of N(2)
We begin by converting the grams of Li and N(2) to moles using the molar mass of each compound.
Li has a molar mass of 6.94 g/mol. *Remember that we can also use the reciprocal of a conversion factor, so we can take the reciprocal of the molar mass, 1 mol/6.94 g, and use this as a conversion factor.* So:
4.0g Li x (1mole Li/ 6.94g Li) = 0.576 moles Li


N(2) has molar mass of 2 x N = 2 x (14.01 g/mol) = 28.02 g/mol
2.5g N(2) x (1mole N(2)/28.02g N(2) = 0.089 moles N(2)

So 4.0g of Li reacting with 2.5g of N(2) is the same thing as 0.576 moles Li reacting with 0.089 moles of N(2).

6 Li + N(2) --> 2 Li(3)N
Now, using the equation, we determine how much product each substance would form using the mole ratios in the equation:
6 moles of Li are needed to produce 2 moles of Li(3)N, so the mole ratio of Li to Li(3)N is 6 to 2, and the mole ratio of Li(3)N to Li is 2 to 6.
So:
0.576 moles Li x (2 mol Li(3)N / 6 mol Li) = 0.192 moles of Li(3)N

1 mole of N(2) is needed to produce 2 moles of Li(3)N, so the mole ratio of the N(2) to Li(3)N is 1 to 2, and the mole ratio of Li(3)N to N(2) is 2 to 1.
So:
.089 moles N(2) x (2 mol Li(3)N / 1 mol N(2) ) = 0.178 moles of Li(3)N

From this we have this information:
4.0g of Li produces 0.192 moles of Li(3)N
2.5g of N(2) produces 0.178 moles of Li(3)N
Since N(2) produces the less amount of Li(3)N, it is the limiting reactant and the Li is the excess reactant. This is true since after 0.178 moles of Li(3)N form, there is not enough N(2) to continue reacting with the Li.

Since the Li is excess reactant and we want to calculate the amount left over, we want to find the amount of Li leftover.
We do this by figuring out how much Li is needed to produce 0.178 moles of Li(3)N, the amount of Li(3)N that is produced.
0.178 moles of Li(3)N x (mole ratio = 6 moles of Li/2 moles Li(3)N) =0.534 moles Li
Then convert back to grams:
0.534 moles Li x (6.94 g/1 mol) = 3.71 g of Li
So 3.71 grams of Li are used to produce 0.178 moles of Li(3)N

Since the original amount of Li is 4.0g,
4.0g - 3.71g = 0.29 g Li
The original amount - amount used up = amount of leftover

So there are 0.29 g Li leftover.