Question

The amount of $\text{BaS}{\text{O}}_{\text{4}}$ formed upon mixing $\text{100ml}$ of $$\text{20}\text{.8 }$$ solution with $\text{50ml}$ of $\text{9}\text{.8 }$ 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.
$\begin{array}{rl}& \text{mass}\text{faction}\text{=}{\displaystyle \frac{\text{w}}{\text{W+w}}}\text{where}\text{;}\text{w=}\text{weight}\text{of}\text{solute}\\ & \text{W=weight}\text{of}\text{solvent}\end{array}$
% 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 $$\text{20}\text{.8 }$$ 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 = ${\displaystyle \frac{20.8}{89.2}}gm$of $$\text{BaC}{\text{l}}_{\text{2}}$$
$\text{100gm}$ of water will contains =$${\displaystyle \frac{\text{20}\text{.8}}{\text{89}\text{.2}}}\times \text{ 100}\text{gm}$$
                                                     = $\text{23}\text{.32gm}$
Thus number of mole of $$\text{BaC}{\text{l}}_{\text{2}}$$ $$\text{mole(n) =}{\displaystyle \frac{\text{mass}}{\text{molar}\text{mass}}}$$ ${\displaystyle \frac{23.32}{208}}=0.112\text{moles}$
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 = ${\displaystyle \frac{9.8}{90.2}}gm$of ${\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}$
$\text{50gm}$ of water will contains =$${\displaystyle \frac{\text{9}\text{.8}}{\text{90}\text{.2}}}\times \text{ 50}\text{gm}$$
                                                     = $\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) =}{\displaystyle \frac{\text{mass}}{\text{molar}\text{mass}}}....(1)$$
  $n={\displaystyle \frac{5.43}{98}}$ $\text{=}\text{0}\text{.0554 moles}$
The overall reaction is the solution is ${\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}\text{+}\text{BaC}{\text{l}}_{\text{2}}\to \text{BaS}{\text{O}}_{\text{4}}\text{+}\text{2HCl}$
Since ${\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}$ is the limiting quantity so it will be the limiting reagent
So, $\text{0}\text{.055 moles}$of ${\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}$ will react with the $\text{0}\text{.055 moles}$of$$\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) =}{\displaystyle \frac{\text{mass}}{\text{molar}\text{mass}}}....(1)$$
$\begin{array}{rl}& \text{0}\text{.055=}{\displaystyle \frac{\text{weight/mass}}{\text{233}}}\\ & \text{weight of BaS}{\text{O}}_4\text{ =12}\text{.8gm}\end{array}$

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.