Question

Chem help please!!!

Consider the reaction:

?Li(s)+?N2(g)—->?Li3N(s)


Calculate the mass of Lithium nitride formed from 33.6524g of nitrogen gas and 58.7032g lithium

Answer 1

The balanced chemical equation for the reaction given is:

`6Li (s) + N_(2)(g) --> 2Li_(3)N (s)`

Moles of
`N_(2) = 33.6524 g * (1 mol)/(28 g) = 1.20187 mol N_(2)`

Moles of Li =
`58.7032 g Li * (1 mol Li)/(6.94 g Li) = 8.4587 mol Li`

Determining the limiting reagent: The reactant that gives least amount of the product will be the limiting reactant.

Mass of
`Li_(3)N from N_(2) = 1.20187 mol N_(2)*(2 mol Li_(3)N)/(1 mol N_(2))*(34.83 g Li_(3)N)/(1 mol Li_(3)N) = 83.7223 g Li_(3)N`

Mass of
`Li_(3)N from Li =`
`8.4587 mol Li *(2 mol Li_(3)N)/(6 mol Li) *(34.83 g Li_(3)N)/(1mol Li_(3)N) = 98.2055 g Li_(3)N`

`N_(2)` will be the limiting reactant and the mass of lithium nitride produced will be dependent on the mass of limiting reactant.

Therefore, mass of lithium nitride produced = 83.7223 g