Question

How many grams of lithium are required to completely react with 58.5 mL of N2 gas measured at STP?
6 Li(s) + N2(g) = 2 Li3N(s)

Answer 1

Answer:- At STP, 22.4 L of gas has 1 mol value. So at STP, the number of moles in 58.5 ml is-

58.5/1000 nN2 = 22.4 - mo/

$\Rightarrow n_{N_{2}}=0.002611\ mol$

from the equation, we see that 1 mol of N2 needs 6 mol of Lithium. So for the above mol of Nitrogen, we need-

$\Rightarrow n_{Li}=0.002611*6\ mol=0.01566\ mol$

Hence the number of grams of Lithium required is-

$\Rightarrow W_{Li}=0.01566*6.94\ gram$

$\boldsymbol{\Rightarrow W_{Li}=0.1087\ gram}$