Lithium reacts with nitrogen gas according to the following reaction: 6Li(s)+N2(g)-->2LiN(s)

What mass of lithium (in g ) is required to react completely with 59.9 mL of N2 gas at STP?

4 answers

Use the ideal gas law..
PV=nRT
and solve for moles of N2

SO....
(1.00atm)(.0599L)=n(0.0821L*atm/mol*K)(273K)
solve for n.

then use "n" moles of N2 and then set up an conversion factor.

so....
___moles of N2 x( 6mol Li/2mol N2) x (MolarMass Li/1mol of Li)

Then that gives you your mass of Lithium required to react.
Hope that helps!
.555 g
.110
0.107