Question

What is the theoretical yield of$L{{i}_{3}}N$ in grams when 12.5 g of Li is heated with 34.1 g of ${{N}_{2}}$?

Answer 1

This problem is based on limiting reagent. It is the reactant that limits the amount of the product formed due to its less quantity. The number of moles of any substance can be given by dividing the given mass upon molar mass. We will take into account the stoichiometric equations to determine the respective masses.

Complete answer:
We have been given two compounds that react together to form the product. Li reacts with ${{N}_{2}}$ to form $L{{i}_{3}}N$, the mass of Li is 12.5 g and that of ${{N}_{2}}$ is 34.1 g. As we have been given the mass of both the reactants, then one of them will be the limiting reagent, and the other will be the excess reagent. The limiting reagent will be consumed completely from its amount. The reaction for the given species is,
$6Li(s)+{{N}_{2}}(g)\to 2L{{i}_{3}}N(s)$
Using stoichiometry we have, the molar mass of Li is 6.941 g/mol, and of ${{N}_{2}}$ is 28.014 g/mol, and that of $L{{i}_{3}}N$ is 3 (6.941 g/mol) + 14.007 g/mol = 34.83 g/mol.
Now, we know that 1 mole that is 6.942 g/mol amount of Li and 28.014 g/mol of ${{N}_{2}}$ produces 34.83 g/mol$L{{i}_{3}}N$, so 12.5 g of lithium (6 moles) and 34.1 g of ${{N}_{2}}$(1 mole) will produce 2 moles $L{{i}_{3}}N$, so individually calculating the yield of $L{{i}_{3}}N$ by dividing the given mass by molar mass, and considering the stoichiometry of the balanced equation and using unitary method we have,
From lithium, the yield of $L{{i}_{3}}N$
$\dfrac{1mol\,Li}{6.941\,g\,Li}\times \dfrac{34.83g\,L{{i}_{3}}N}{1mol\,L{{i}_{3}}N}\times \dfrac{2\,mol\,L{{i}_{3}}N}{6mol\,Li}\times 12.5g\,Li$ = 20.9 g $L{{i}_{3}}N$.
From nitrogen gas,
$\dfrac{1mol\,{{N}_{2}}}{28.01\,g\,{{N}_{2}}}\times \dfrac{34.83g\,L{{i}_{3}}N}{1mol\,L{{i}_{3}}N}\times \dfrac{2\,mol\,L{{i}_{3}}N}{1mol\,{{N}_{2}}}\times 34.1g\,{{N}_{2}}$= 84.8 g $L{{i}_{3}}N$
As lithium produces less $L{{i}_{3}}N$, so it is the limiting reagent and nitrogen gas is the excess reagent, so lithium will be completely consumed and limit the production of $L{{i}_{3}}N$ to 20.9 g.
Hence, the theoretical yield of $L{{i}_{3}}N$ is 20.9 grams.

Note:
We can identify any problem to be of limiting reagent if we have been given the mass or amount of both the reactants in the problem. These problems require stoichiometric calculations that involve a balanced chemical equation to give the exact number of the quantities involved. Any mistake in the balancing would yield incorrect results.