How many grams of lithium are required to completely react with 59.3 ml of n2 gas at stp?

At STP, that is standard temperature and pressure conditions, 1 mol of any gas occupies a volume of 22.4 L

the balanced equation for the reaction between Li and N₂;

6Li + N₂ - - > 2Li₃N

stoichiometry of Li to N₂ is 6:1

since 22.4 L is occupied by 1 mol

the number of moles occupying 0.0593 mL is - 1/22.4 x 0.0593 = 0.00265 mol

the number of N₂ moles reacted - 0.00265 mol

if 1 mol of N₂ reacts with 6 mol of Li

then 0.00265 mol of N₂ reacts with - 6 x 0.00265 = 0.0159 mol

the mass of Li reacted is 0.0159 mol x 7 g/mol = 0.111 g

mass of Li required - 0.111 g