What is the balanced equation? How many moles of chlorine were available to react? ....?

a. Write a balanced equation for this reaction.

Balanced equation:
2Na(s) + Cl2(g) ---> 2NaCl(s)

Since both reactants/reagents are involved in making product,
once one reactant runs out, the other has nothing to react with.
The reagent that runs out first, is the limiting reagent.
The one that is left with nothing to react with, is the excess reagent.

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b. How many moles of sodium metal were available to react?
5.0 g Na x (1 mol Na / 22.990 g Na)
= 0.21749 mol Na

Finding mol NaCl, starting from g Na reagent:

5.0 g Na
x (1 mol Na / 22.990 g Na) - from atomic weight
x (2 mol NaCl / 2 mol Na) - from balanced equation
= 0.21749 mol NaCl could be produced.

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c. How many moles of chlorine gas were available to react?
0.08533 mol Cl2

Assuming Cl2 behaves like an ideal gas:
PV = nRT

The moles of Cl2:

Solving for n:
n = PV / RT
where,
P = 780 mmHg
V = 2.0 L
R = 62.3638 mmHg·L/mol·K
T = 20ºC + 273.15 = 293.15 K

Substituting numbers:
n = (780 mmHg x 2.0 L) / [(62.3638 mmHg·L/mol·K) x 293.15 K]
= 0.08533 mol Cl2

0.08533 mol Cl2
x (2 mol NaCl / 1 mol Cl2) - from balanced equation
= 0.17066 mol NaCl could be produced.

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d. Which substance was the limiting reagent (of which was there less)?
Cl2 is the limiting reagent.

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e. What mass of sodium chloride was produced?

0.17066 mol NaCl x (58.443 g NaCl / 1 mol NaCl)
= 9.97388 g NaCl

9.97 g NaCl (at e sig. figs.)

10. g NaCl (at 2 sig. figs.)