Part 1. 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.

F2 + 2NaCl → Cl2 + 2NaF

Part 2. Explain how you would determine the mass of sodium chloride that can react with the same volume of fluorine gas at STP.

1. 91.728g of NaCl

2. 62.712g of NaCl

Explanation:

Part 1:

We obtained the following information from the question:

V = 12L

P = 1.5atm

T = 280K

R = 0.082atm. L/K / mol

n

PV = nRT

n = PV / RT

n = (12x1.5) / (0.082x280)

n = 0.784mol

Converting the number of mole of F2 obtained to mass, we have:

MM of F2 = 2 x 19 = 38g/mol

Mass of F2 = 0.784 x 38 = 29.792g

F2 + 2NaCl → Cl2 + 2NaF

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

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

From the equation,

38g of F2 reacted with 117g of NaCl.

Therefore

29.792g of F2 will react with = (29.792 x 117) / 38 = 91.728g of NaCl

Part 2:

1mole of a gas occupy 22.4L at stp. This implies that 1mole of F2 also occupy 22.4L at stp. Now, we must determine how many moles of F2 will occupy 12L as given from the question. This can be achieved by doing the following:

22.4L = 1mol

12L = 12/22.4 = 0.536mol

Now let us convert this mole of F2 (i. e 0.536mol) to grams.

Mass of F2 = 0.536 x 38 = 20.368g

Now let us consider the equation.

From the equation,

38g of F2 reacted with 117g of NaCl.

Therefore,

20.368g of F2 will react with = (20.368x117) / 38 = 62.712g of NaCl