Gas Law Stoichiometry Question?

A.
Initial volume of H₂ = 15.00 L
Initial volume of Cl₂ = 10.00 L

Equation for the reaction: H₂ + Cl₂ → 2HCl
At constant temperature and pressure, 1 L of H₂ reacts with 1 L of Cl₂ to give 2 L of HCl.

When 10 L of Cl₂ completely reacts, H₂ needed = 10.00 L < 15.00 L
Hence, H₂ is in excess, and Cl₂ is the limiting reagent.

Volume of HCl produced = 10 × 2 L = 20 L

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B.
Refer to part A.
Cl₂ would be the limiting reagent.

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C.
For the excess H₂:
Pressure, P = 0.955 atm
Volume, V = (15.00 - 10.00) L = 5.00 L
Molar mass, M = 1.0×2 g/mol = 2.0 g/mol
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = (273 + 26) K = 299 K

PV = nRT   and   n = m/M
Then, PV = (m/M)RT
Hence, m = PVM/(RT)

Mass of O₂, m = 0.955 × 5.00 × 2.0 / (0.08206 × 299) g = 0.389 g

Hydrogen + Chlorine ➜ Hydrogen Chloride
Cl₂ + H₂ = 2HCl
molecular mass
Cl₂ = 2•35.45 = 70.9
H₂ = 2.01
2HCl = 72.91

1 mol of Cl₂ + 1 mole of H₂ = 2 moles of HCl
or
70.9 g of Cl₂ + 2.01 g of H₂ = 72.91 g of HCl


for perfect gases, number of liters, at same pressure and temp, is proportional to number of moles. 

so
1 liter of Cl₂ + 1 liter of H₂ = 2 liter of HCl

You have 15.00 liters of hydrogen reacting with 10.00 liters of chlorine.
correct ratio is 1:1 so H₂ is in excess, as only 10 L is needed. given that, 10 L of Cl₂ will produce 20 L of HCl

there would be 5 L of H₂ remaining.

Ideal gas law
PV = nRT
n = number of moles
R = gas constant = 0.08206 (atm∙L)/(mol∙K)
                   8.314 L∙kPa/mol∙K
                   0.08314 L∙bar/mol∙K
                   62.36 L∙torr/mol∙K
                   62.36 L∙mmHg/mol∙K
T = temperature in kelvins
P = absolute pressure in atm, kPa, bar, torr
V = volume in liters

26 degrees C and 0.955 atm

n = PV/RT = (0.955)(5) / (0.08206)(273.16+26) moles

and H₂ is 2 g/mol, so 
number of grams is 2 g/mol x n 
   = 2(0.955)(5) / (0.08206)(273.16+26)
   = (9.55) / (0.08206)(299.16)
(surely you have a calculator...)