How many grams of sodium metal are needed to react completely with 25.8 liters of chlorine gas at 293 kelvin and 1.30 atmospheres? ideal gas law after stoichiometry

First, we need to get the moles of Cl2 ( chlorine)

we can use the ideal gas formula to get n ( moles of gas) :

PV = nRT

when p is the pressure = 1.3 atm 

and V is the volume = 25.8 L 

R is the constant of the ideal gas = 8.314

T temperature in Kelvin = 293 

∴ n = PV / RT

      = 1.3 * 25.8 / 8.314* 293

      = 0.0138 moles 

according to the reaction equation:

2Na(s) +Cl2 → 2 NaCl 

we can see from the balanced equation that the molar ration between Na & Cl2 = 2: 1 

∴ moles of Na = moles of Cl2 / 2
                        = 0.0138 / 2
                        = 0.0069 moles 

now, we have the moles of Na and by using the molar mass of Na = 22.99 g/mol 

∴ Mass of Na = moles of Na * molar mass  

                       = 0.0069 moles * 22.99

                      = 0.1586 g