Question

Lithium reacts with nitrogen gas according to the following reaction: 6Li(s)+N2(g)?2Li3N(s) What mass of lithium is required to react completely with 55.4 mL of N2 gas at STP? Express your answer in grams.

Answer 1

volume = 55.4 mL = 55.4 x 10^-3 L

STP : P = 1 atm

          T = 273 K

P V = n R T

1 x 55.4 x10^-3 = n x 0.0821 x 273

n = 2.47 x 10^-3

6 mole Li --------------------> 1 mole N2

X mole Li ---------------------> 2.47 x 10^-3 mol N2

X = 2.47 x 10^-3 x 6 / 1

X = 0.0148

moles of Li = 0.0148

mass of Li = 0.0148 x 6.94

                  = 0.103 g

mass of Li = 0.103 g