Calculate the mass of solid sodium that's needed to fully react with 4.0 l of chlorine gas at stp.
_na ₍s₎+_cl₂₍g₎ →_nacl₍aq₎

8.28g Na

Explanation: use ideal gas law

PV= nRT

Solve for moles of Cl2

n= PV/ RT

Substitute:

= 1 atm x 4.0 L / 0.08205 L.atm/ mol. K x 273 K

= 0.18 moles Cl2

Do stoichiometry to solve for m of Na

2 Na + Cl2 => 2 NaCl2

=0.18 moles Cl2 x 2 mol Na/ 1 mol Cl2 x 23g Na / 1 mol Na

= 8.28 g Na.

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.  

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