Question

For ${H_2} + C{l_2} \to 2HCl$
if \[2.02g\] of ${H_2}$ and \[35.45g\] of $C{l_2}$ are allowed to react, what is the limiting reactant and what mass of $HCl$ Will it be formed?
(A) ${H_2},72.92g$
(B) $C{l_2},36.46g$
(C) ${H_2},36.46g$
(D) $C{l_2},18.23g$

Answer 1

Hint:A reactant that gets completely consumed and limits the formation of a product is known as limiting reagent. It also identifies the excess quantity of the other reagent which does not get completely consumed.

Complete step by step answer:
When during a chemical reaction, the reactants are not present in the amount as required by the balanced chemical equation, then one reactant gets consumed and the excess of the other reactant remains due to a cease of reaction.
In this question we are provided with the equation:
${H_2} + C{l_2} \to 2HCl$
The mass of ${H_2} = 2.02g$
Mass of $C{l_2} = 35.45g$
For finding the limiting reagent we must model each reaction.
Thus, to find the number of moles we will use the formula.
Number of moles $(n)$ $ = $ $\dfrac{m}{M}$
Where $m$ is given mass and $M$ is given molar mass.
For Hydrogen ${H_2}$
$
  n = \dfrac{m}{M} \\
    \\
 $
$
   \Rightarrow n = \dfrac{{2.02}}{{2.02}} \\
    \\
 $
$ \Rightarrow 1mole$

So, number of moles of hydrogen is $1$ .
For Chlorine $C{l_2}$
$
  n = \dfrac{m}{M} \\
    \\
 $
$
   \Rightarrow n = \dfrac{{35.45}}{{70.9}} \\
    \\
 $
$ \Rightarrow n = 0.5moles$
So, number of moles of Chlorine is $0.5$
 According to the equation, one mole of ${H_2}$ reacts with one mole of $C{l_2}$ to give two moles of $HCl$.
We also now know the $C{l_2}$ is the limiting reagent because there are only \[0.5{\text{ }}moles\] of $C{l_2}$.
So, amount that reacts with $0.5moles$$C{l_2}$
$ \Rightarrow 0.5 \times \dfrac{1}{1}$
$ \Rightarrow 0.5moles$
Therefore, when $0.5moles$ of ${H_2}$ and $0.5moles$ of $C{l_2}$ $
   = 0.5 + 0.5 \\
    \\
 $
$ \Rightarrow 1mole$
So, one mole $HCl$ is formed.
Also, the mass of $1mole$ ,$HCl$ will be $ = 1.01 + 35.45$
mass of $HCl = 36.46g$
Hence, \[36.46g\] of $HCl$ is formed.
Hence, the correct option is (B) $C{l_2},36.46g$.

Note:
After the complete usage of limiting reagent, the proceeding of the reaction stops because the limiting reactant gets exhausted. Therefore, the products formation would depend majorly upon the limiting reagent and less on the reactant in excess.