1.Chlorine gas can be produced commercially

Question

by passing an electric current through a concentrated
solution of sodium chloride (brine).
The unbalanced equation is
NaCl(aq) + H2O(ℓ) →
NaOH(aq) + Cl2(g) + H2(g).
If the brine contains 163 g of NaCl, how
much Cl2 can be produced?
Answer in units of mol.
This is what I did
(163 g NaCl) / (58.4430 g NaCl/mol) x (1/2) = 1.39 mol Cl2

(Part 2 of 2)How much H2 can be produced?
Answer in units of mol.
I don't know how to do the second questions since they didn't give how much h2 weighs
Please help me set up the problem

2.Phosphorous trichloride (PCl3) is produced
from the reaction of white phosphorous (P4)
and chlorine:
P4(s) + 6 Cl2(g) → 4 PCl3(g).
A sample of PCl3 of mass 300.1 g was collected
from the reaction of 72.55 g of P4 with excess
chlorine. What is the percentage yield of the
reaction?
Answer in units of %.

Answer 1

I think for part 1 they want grams. Your 1.39 mols Cl2 is right; just multiply by molar mass Cl2.

part 2. H2 done the same way you did #1.
mols NaCl = 163/58.44 = ?
mols H2 = 1/2 that from the balanced equation.
grams H2 = mols H2 x molar mass H2 which is 2 g/mol.

#2. P4(s) + 6 Cl2(g) → 4 PCl3(g).
mols P4 = 72.55/molar mass P4 = ?
Convert mols P4 to mols PCl3 using the coefficients in the balanced equation. That's ?mols P4 x (4 mols PCl3/1 mol P4) = ? mols PCl3.
Then g PCl3 = mols PCl3 x molar mass PCl3. This is the theoretical yield(TY); ie., as if it were 100%. The actual yield (AY) is stated in the problem as 72.55.
Then % yield = (AY/TY)*100 = ?