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 :

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 gas by using ideal gas equation.

where,

P = pressure of gas = 1.50 atm

V = volume of gas = 15.0 L

T = temperature of gas = 280 K

n = number of moles of gas = ?

R = gas constant =

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

Now we have to calculate the moles of NaCl.

The balanced chemical reaction is,

From the balanced reaction, we conclude that

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

So, 0.979 mole of gas react with moles of NaCl

Now we have to calculate the mass of NaCl.

The mass of NaCl reacted is, 114.543 grams.

For part 2 :

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

where,

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

V = volume of gas = 15.0 L

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

n = number of moles of gas = ?

R = gas constant =

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

Now we have to calculate the moles of NaCl.

The balanced chemical reaction is,

From the balanced reaction, we conclude that

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

So, 0.669 mole of gas react with moles of NaCl

Now we have to calculate the mass of NaCl.

The mass of NaCl reacted is, 78.273 grams.

is there multiple choice answers?

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.