The mass of NaCl that reacted with 12 lit of F2 at 280 K and 1.50 atm= 0.027 gm.
the mass of sodium chloride that can react with the same volume of fluorine gas at STP = 0.01 gm.
Given,
F2 + 2NaCl → Cl2 + 2NaF
PART-1:
So, from the given reaction we can say that 2 moles of NaCl is required for 1 mole of F2.
By using combined gas law,
[tex]\frac{P1V1}{n1T1} =\frac{P2V2}{n2T2}[/tex]
here , P1 =P2
and moles of NaCl is twice as pf F2
∴n2=2n1
and also T1 = T2
by simplifying the reaction we get..
[tex]\frac{V1}{n1} = \frac{V2}{2n1}[/tex]
V2= 2 × V1
∴V2 = 2 ×12
   = 24 Lit.
from ideal gas law ,
we can find the moles of NaCl,
PV= nRT
n=[tex]\frac{PV}{RT}[/tex]
n= 1.5 × 24 ÷( 0.082 × 280)
∴ n= 1.57
mole = mass / molecular mass
so, molecular mass of NaCl = 58.5
∴ mass of NaCl used = 1.57/58.5
                 = 0.027 gm
PART -2:
At, STP
T= 273 K
P = 1atm
V= 12 lit
n= ?
using ideal gas law we can determine the number of moles
PV= nRT
n = 1 ×12 ÷(0.082 × 273)
n= 0. 54
∴ mass of NaCl = 0.54 / 58.5
             = 0.01 g
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