Question

How many moles of $ NaCl $ will react completely with $ 18.5L $ $ F_2 $ gas at $ 300.0K $ and $ 1.00mol $ $ atm $ ?
(A) $ 0.75mol $ $ NaCl $
(B) $ 1.33mol $ $ NaCl $
(C) $ 1.50mol $ $ NaCl $
(D) $ 2.66mol $ $ NaCl $

Answer 1

Hint: For this kind of question we will first have to write balanced equation through which we will get a relation between two quantities of molecules , after that by using basic chemistry formula in which we can use known quantities and find out unknown or required quantity and then by putting the value we got in the previous relation we got through reaction we will get the answer.

Complete step by step solution:
To understand the question better, we will first write down the reaction which is taking place;
 $ 2NaCl + {F_2} \to 2NaF + C{l_2} $
The above equation is balanced and from that we know $ 2mol $ of $ NaCl $ corresponds to $ 1mol $ of $ {F_2} $
Now, for calculation of moles of $ NaCl $ reacting with $ F_2 $ , we will first calculate the number of moles of $ F_2 $ gas.
By using basic ideal gas law;
 $ pV = nRT $
Where $ p = $ pressure
 $ V = $ volume
 $ n = $ number of moles
 $ R = $ gas constant
 $ T = $ temperature
Now we want to know number of moles every thing is already given and we know $ R = 0.08206L $
 $ n = \dfrac{{pV}}{{RT}} = \dfrac{{1 atm \times 18.5L}}{
  0.08206L.atm/K.mol \times 300K \\
   = 0.751mol } $
And as we know one mole of $ F_2 $ gas corresponds to two moles of $ NaCl $
Moles of $ NaCl $
   $ = 2 \times 0.751 \\
   = 1.50moles.$
Therefore, 1.50 moles of $ NaCl $ will react completely with $ 18.5L $ $ F_2 $ gas at $ 300.0K $ and $ 1.00mol $ $ atm $. So, Option (C) is correct.

Note:
The gas constant which we have used in the above example is defined as the product of volume and pressure. It is an important physical constant used in many fundamental equations. We can express it in many ways with different unit systems.

Values of R	Units
$ 8.31 $	$ J/mol.K $
$ 1.98 $	$ C/mol.K $
$ 8.31 $	$ {m^3}.Pa/mol.K $
$ 0.0821 $	$ L.atm/mol.K $
$ 1.98 \times {10^{ - 3}} $	$ K.Cal/mol.K $
$ 62.36 $	$ L.torr/mol.K $