Part 1. A chemist reacted 15.0 liters of F2 gas with NaCl in the laboratory to form Cl2 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.

Question

Cassidy Tapia

Chemistry

26 October, 07:20

Part 1. A chemist reacted 15.0 liters of F2 gas with NaCl in the laboratory to form Cl2 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.

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Answers (1)

Know the Answer?

Ali Silva · Answer 1

1 - 114.54 g.

2 - 78.32 g.

Explanation:

Q1:

The balanced equation of the reaction is:

F₂ + 2NaCl → Cl₂ + 2NaF.

It is clear that 1.0 mole of F₂ reacts with 2.0 moles of NaCl to produce 1.0 mole of Cl₂ and 2.0 moles of NaF. We can calculate the no. of moles of F₂ (15.0 liters at 280.0 K and 1.50 atm) using the ideal gas law:

PV = nRT,

where, P is the pressure of the gas (P = 1.0 atm),

V is the volume of the gas (V = 15.0 L),

n is the no. of moles of the gas (n = ? mole),

R is the general gas constant (R = 0.082 L. atm/mol. K),

T is the temperature of the gas (T = 280.0 K).

∴ n = PV/RT = (1.5 atm) (15.0 L) / (0.082 L/atm/mol. K) (280.0 K) = 0.98 mol.

We can get the no. of moles of NaCl reacted with 0.98 mol of F₂.

Using cross multiplication:

1.0 mole of F₂ reacts with → 2.0 moles of NaCl, from the stichiometry.

0.98 mole of F₂ reacts with →? moles of NaCl.

∴ The no, of moles of reacted NaCl = (2.0 mole) (0.98 mole) / (1.0 mole) = 1.96 mol.

Now, we can get the mass of NaCl that reacted with F₂ at 280.0 K and 1.50 atm:

∴ The mass of NaCl = n x molar mass of NaCl = (1.96 mol) (58.44 g/mol) = 114.54 g.

Q2:

STP conditions means that P = 1.0 atm and T = 273.0 K. We can calculate the no. of moles of F₂ (15.0 liters at 273.0 K and 1.0 atm) using the ideal gas law: PV = nRT,

∴ n = PV/RT = (1.0 atm) (15.0 L) / (0.082 L/atm/mol. K) (273.0 K) = 0.67 mol.

We can get the no. of moles of NaCl reacted with 0.67 mol of F₂.

Using cross multiplication:

1.0 mole of F₂ reacts with → 2.0 moles of NaCl, from the stichiometry.

0.67 mole of F₂ reacts with →? moles of NaCl.

∴ The no, of moles of reacted NaCl = (2.0 mole) (0.67 mole) / (1.0 mole) = 1.34 mol.

Now, we can get the mass of NaCl that reacted with F₂ at 280.0 K and 1.50 atm:

∴ The mass of NaCl = n x molar mass of NaCl = (1.34 mol) (58.44 g/mol) = 78.32 g.