Question

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.

Answer 1

Answer :

For part 1 : The mass of NaCl reacted is, 114.543 grams.

For part 1 : The mass of NaCl reacted is, 78.273 grams.

Explanation :

For part 1 :

First we have to calculate the volume of for mole of $F_2$ gas by using ideal gas equation.

$PV=nRT$

where,

P = pressure of $F_2$ gas = 1.50 atm

V = volume of $F_2$ gas = 15.0 L

T = temperature of $F_2$ gas = 280 K

n = number of moles of $F_2$ gas = ?

R = gas constant = $0.0821L.atm/mole.K$

Now put all the given values in the ideal gas equation, we get:

$(1.50atm)\times (15.0L)=n\times (0.0821L.atm/mole.K)\times (280K)$

$n=0.979mole$

Now we have to calculate the moles of NaCl.

The balanced chemical reaction is,

$F_2+2NaCl\rightarrow Cl_2+2NaF$

From the balanced reaction, we conclude that

As, 1 mole of $F_2$ gas react with 2 moles of NaCl

So, 0.979 mole of $F_2$ gas react with $2\times 0.979=1.958$ moles of NaCl

Now we have to calculate the mass of NaCl.

$\text{Mass of }NaCl=\text{Moles of }NaCl\times \text{Molar mass of }NaCl$

$\text{Mass of }NaCl=1.958mole\times 58.5g/mole=114.543g$

The mass of NaCl reacted is, 114.543 grams.

For part 2 :

First we have to calculate the volume of for mole of $F_2$ gas by using ideal gas equation.

$PV=nRT$

where,

P = pressure of $F_2$ gas = 1 atm (at STP)

V = volume of $F_2$ gas = 15.0 L

T = temperature of $F_2$ gas = 273 K  (at STP)

n = number of moles of $F_2$ gas = ?

R = gas constant = $0.0821L.atm/mole.K$

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (15.0L)=n\times (0.0821L.atm/mole.K)\times (273K)$

$n=0.669mole$

Now we have to calculate the moles of NaCl.

The balanced chemical reaction is,

$F_2+2NaCl\rightarrow Cl_2+2NaF$

From the balanced reaction, we conclude that

As, 1 mole of $F_2$ gas react with 2 moles of NaCl

So, 0.669 mole of $F_2$ gas react with $2\times 0.669=1.338$ moles of NaCl

Now we have to calculate the mass of NaCl.

$\text{Mass of }NaCl=\text{Moles of }NaCl\times \text{Molar mass of }NaCl$

$\text{Mass of }NaCl=1.338mole\times 58.5g/mole=78.273g$

The mass of NaCl reacted is, 78.273 grams.

Answer 2

Okey so 1st find the moles of F2 which is :

n = 15/22.4 = 0.667mol

Then according to stoichiometric values, F2 and NaCl, they are to a ratio of 1 : 2

So take the twice value of the calculated moles.

Use the moles to find the mass using the equation :

m = n×M

Where m is the mass

n is moles

M is molar mass (of NaCl)

Therefore m = 1.334×57.5

(given the molar mass of Na 22gmol^-1 and CL 35.5gmol^-1)

m = 76.705g