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Question

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.

Answer 1

First, let's write down the balanced equation for the reaction taking place.

F₂(g) + 2NaCl ==> Cl₂(g) + 2NaF .. balanced equation

Ideal Gas Law:

PV = nRT and we will use this equation to first find the moles of F2 in the above reaction. From that we will calculate the moles of NaCl needed and finally, convert moles NaCl to grams NaCl.

P = pressure = 1.50 atm

V = volume in liters = 12.0 L

n = moles F₂ = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 280K

Solving for n (moles):

n = PV/RT = (1.50)(12.0) / (0.0821)(280)

n = 0.7830 mols F₂

Use the stoichiometry of the balanced equation to find moles of NaCl:

0.7830 mols F₂ x 2 mols NaCl / mole F₂ = 1.566 mols NaCl

Use molar mass of NaCl (58.44 g/mol) to find grams NaCl:

1.566 mols NaCl x 58.44 g / mol = 91.5 g NaCl (3 sig.figs.)

At STP (standard temperature and pressure), 1 mole of any ideal gas = 22.4 liters. Using this relationship, we can determine the mass of NaCl needed under STP conditions.

In part 1, the volume of F₂ gas is 12.0 L. From this we can determine mols of F₂ gas

12.0 L F₂ x 1 mol / 22.4 L = 0.5357 mols F₂

Now, calculate mols NaCl needed:

0.5357 mols F₂ x 2 mols NaCl / mol F₂ = 1.071 mols NaCl

Finally convert to grams of NaCl:

1.071 mols NaCl x 58.44 g / mol = 62.6 g NaCl (3 sig.figs.)