Question

The amount of $\text{BaS}{{\text{O}}_{\text{4}}}$ formed upon mixing $\text{100ml}$ of \[\text{20}\text{.8 }\!\!%\!\!\text{ }\,\text{BaC}{{\text{l}}_{\text{2}}}\] solution with $\text{50ml}$ of $\text{9}\text{.8 }\!\!%\!\!\text{ }\,{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ solution will be:(given the molecular weight of $\text{Ba =137,}\,\text{Cl =35}\text{.5,}\,\text{S =32,}\,\text{H =1}\,\text{and}\,\text{O =16gm/mol}$)
(A) $\text{23}\text{.3gm}$
(B) $\text{11}\text{.65gm}$
(C) $\text{30}\text{.6gm}$
(D) $\text{33}\text{.2gm}$

Answer 1

Hint: Percentage by weight (w/W) is the mass of solute present in the given mass of solution, this is known as the weight fraction of solute.
$\begin{align}
  & \text{mass}\,\text{faction}\,\text{=}\dfrac{\text{w}}{\text{W+w}}\,\,\text{where}\,\text{;}\,\text{w=}\,\text{weight}\,\text{of}\,\text{solute} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{W=weight}\,\text{of}\,\text{solvent} \\
\end{align}$
% concentration (w/W) is generally defined x gram of solute is present in the 100gm of solution.

Complete answer:
To calculate the weight of $\text{BaS}{{\text{O}}_{\text{4}}}$ firstly we will calculate the number of moles of each reactant in the solution.
For \[\text{20}\text{.8 }\!\!%\!\!\text{ }\,\text{BaC}{{\text{l}}_{\text{2}}}\] means $\text{20}\text{.8gm}$ of \[\text{BaC}{{\text{l}}_{\text{2}}}\] is dissolved in $89.\text{2gm}$ of water.
Since $\text{100ml =100gm}$ so, amount of \[\text{BaC}{{\text{l}}_{\text{2}}}\] present in the 100gm of water solution will be-
From unitary method
if \[\text{89}\text{.2gm}\]of contains = $\text{20}\text{.8gm}$ of \[\text{BaC}{{\text{l}}_{\text{2}}}\]
$\text{1gm}$of water will contain = $\dfrac{20.8}{89.2}gm$of \[\text{BaC}{{\text{l}}_{\text{2}}}\]
$\text{100gm}$ of water will contains =\[\dfrac{\text{20}\text{.8}}{\text{89}\text{.2}}\text{ }\!\!\times\!\!\text{ 100}\,\text{gm}\]
                                                     = $\text{23}\text{.32gm}$
Thus number of mole of \[\text{BaC}{{\text{l}}_{\text{2}}}\] \[\text{mole(n) =}\,\dfrac{\text{mass}}{\text{molar}\,\text{mass}}\] $\dfrac{23.32}{208}=\,\,0.112\,\text{moles}$
In the same way we will calculate the moles for${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$-
9.8% ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$means $\text{9}\text{.8gm}$ of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ is dissolved in $\text{90}\text{.2gm}$ of water.
Since $\text{50ml = 50gm}$ so, amount of${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$present in the 50gm of water solution
From unitary method
if \[\text{90}\text{.2gm}\] of contains = $\text{9}\text{.8gm}$ of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$
$\text{1gm}$of water will contain = $\dfrac{9.8}{90.2}gm$of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$
$\text{50gm}$ of water will contains =\[\dfrac{\text{9}\text{.8}}{\text{90}\text{.2}}\text{ }\!\!\times\!\!\text{ 50}\,\text{gm}\]
                                                     = $\text{5}\text{.43gm}$${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$
Thus number of mole of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ \[\text{mole(n) =}\,\dfrac{\text{mass}}{\text{molar}\,\text{mass}}....(1)\]
  $n=\dfrac{5.43}{98}$ $\text{=}\,\text{0}\text{.0554 moles}$
The overall reaction is the solution is ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\,\text{+}\,\text{BaC}{{\text{l}}_{\text{2}}}\,\,\to \,\,\text{BaS}{{\text{O}}_{\text{4}}}\,\text{+}\,\text{2HCl}$
Since ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ is the limiting quantity so it will be the limiting reagent
So, $\text{0}\text{.055 moles}$of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ will react with the $\text{0}\text{.055 moles}$of\[\text{BaC}{{\text{l}}_{\text{2}}}\] and it will form $\text{0}\text{.055 moles}$ of $\text{BaS}{{\text{O}}_{\text{4}}}$.To calculate the weight of $\text{BaS}{{\text{O}}_{\text{4}}}$ we will apply mole formula. So, after putting the value of the number of moles and molar mass of $\text{BaS}{{\text{O}}_{\text{4}}}$ in the e.q…. (1)
\[\text{mole(n) =}\,\dfrac{\text{mass}}{\text{molar}\,\text{mass}}....(1)\]
$\begin{align}
  & \text{0}\text{.055=}\dfrac{\text{weight/mass}}{\text{233}} \\
 & \text{weight of BaS}{{\text{O}}_{4}}\text{ =12}\text{.8gm} \\
\end{align}$

So, option (B) will be the correct answer.

Note:
Limiting reagent is the reagent that is entirely consumed when a reaction goes to completion. If mass of the two or more reactants are given, the amount of product formed depends upon the amount of limiting reagent.