A chemist reacted 12.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equation to determine the mass of NaCl that reacted with F2 at 280. K and 1.50 atm

Answer: 91.73g of NaCl

Explanation:

First, we solve for the number of moles of F2 using the ideal gas equation

V = 12L

P = 1.5 atm

T = 280K

R = 0.082atm. L/mol/K

n = ?

PV = nRT

n = PV / RT

n = (1.5x12) / (0.082x280)

n = 0.784mol

Next, we convert this mole (i. e 0.784mol) of F2 to mass

MM of F2 = 19x2 = 38g/mol

Mass conc of F2 = n x MM

= 0.784 x 38 = 29.792g

Equation for the reaction is given below

F2 + 2NaCl - > 2NaF + Cl2

Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol

Mass conc. of NaCl from the equation = 2 x 58.5 = 117g

Next, we find the mass of NaCl that reacted with 29.792g of F2.

From the equation,

38g of F2 redacted with 117g of NaCl.

Therefore, 29.792g of F2 will react with Xg of NaCl i. e

Xg of NaCl = (29.792 x 117) / 38

= 91.73g

Therefore, 91.73g of NaCl reacted with f2