Question

\[\text{10g}\] of a mixture of $\text{BaO}$and $\text{CaO}$ requires \[\text{100c}{{\text{m}}^{\text{3}}}\] \[\text{2}\text{.5M HCl}\] to react completely. The percentage of calcium oxide in the mixture is approximate: (Given: molar mass of\[\text{BaO = 153}\])
A) \[55.1{\scriptstyle{}^{0}/{}_{0}}\]
B) \[47.4{\scriptstyle{}^{0}/{}_{0}}\]
C) \[52.6{\scriptstyle{}^{0}/{}_{0}}\]
D)\[44.9{\scriptstyle{}^{0}/{}_{0}}\]

Answer 1

Hint: To solve this type of question consider that at the equivalence weight of acid is always equal to the base after the reaction. Thus the amount of $\text{CaO}$ and $\text{BaO}$ reacting is equal to the amount acid required for complete reaction. Use the molarity relation to find out the equivalent masses of solution \[\text{Equivalents of }\!\!~\!\!\text{ CaO }\!\!~\!\!\text{ +Equivalents of }\!\!~\!\!\text{ BaO }\!\!~\!\!\text{ = }\!\!~\!\!\text{ Equivalents of }\!\!~\!\!\text{ HCl}\]

Complete answer:
Let the mass of $\text{CaO}$ reacting from the mixture of $\text{CaO+BaO}$ being the $\text{x gram}$
Since we know that the total $\text{10 gram}$ of the mixture undergoes the reaction with $\text{HCl}$ , therefore, the amount of $\text{CaO}$ undergoes the mixture $\text{=(10-x) gram}$
To find out the percentage of oxide in the mixture, let's first start with calculating the equivalent of $\text{CaO}$ and $\text{BaO}$

1) Equivalent mass of$\text{BaO}$:
The mass of $\text{CaO}$$x\text{ g}$
Therefore, the mass of $\text{BaO=}(10-x)\text{ g}$
\[\text{Equivalence mass of BaO =}\dfrac{\text{molar mass of BaO}}{\text{2}}\]
$\begin{align}
  & =\dfrac{153}{2} \\
 & =76.5 \\
\end{align}$
(Since 2 is the valence number for $\text{BaO}$ and 153 is a molar mass of $\text{BaO}$)
The number of moles $\text{BaO}$is given as,
$\text{No}\text{. of moles of BaO=}\dfrac{\text{10-x}}{\text{76}\text{.5}}$
2) Equivalent mass of $\text{CaO}$:
The mass of $\text{CaO}$$x\text{ g}$
\[\text{Equivalence mass of CaO=}\dfrac{\text{molar mass of CaO}}{\text{2}}\]
$\begin{align}
  & =\dfrac{56}{2} \\
 & =28 \\
\end{align}$
(Since 2 is the valence number for $\text{CaO}$ and 56 is a molar mass of$\text{CaO}$)
The number of moles $\text{CaO}$is given as,
$\text{No}\text{. of moles of CaO=}\dfrac{\text{x}}{\text{28}}$
\[\text{Equivalents of }\!\!~\!\!\text{ CaO }\!\!~\!\!\text{ +Equivalents of }\!\!~\!\!\text{ BaO }\!\!~\!\!\text{ = }\!\!~\!\!\text{ Equivalents of }\!\!~\!\!\text{ HCl}\]
Let's find out the equivalent weight of \[\text{HCl}\],
$\text{Molarity =}\dfrac{\text{moles of solute}}{\text{Volume of solution (in L)}}$
We are given with molarity of \[\text{HCl=2}\text{.5M}\]
$\text{The volume of required=100 c}{{\text{m}}^{\text{3}}}\text{=}\dfrac{\text{100}}{\text{1000}}\text{L=0}\text{.1L}$
Now substitute all values.
$\text{Molarity =}\dfrac{\text{moles of solute}}{\text{Volume of solution (in L)}}$
$\text{2}\text{.5 M=}\dfrac{\text{moles of solute}}{\text{0}\text{.1 L}}$
Or $\text{moles of solute=}\left( \text{0}\text{.1L} \right)\text{ }\!\!\times\!\!\text{ 2}\text{.5 M}$
Or $\text{moles of solute=0}\text{.25M}$
Let's use the formula to find the values of
\[\text{Equivalents of }\!\!~\!\!\text{ CaO }\!\!~\!\!\text{ +Equivalents of }\!\!~\!\!\text{ BaO }\!\!~\!\!\text{ = }\!\!~\!\!\text{ Equivalents of }\!\!~\!\!\text{ HCl}\]
$0.25=\dfrac{x}{28}+\dfrac{10-x}{76.5}$
Or $0.25\times 28\times 76.5=76.5x+28(10-x)$
Or $255.5=48.5x$
Or$x=\dfrac{255.5}{48.5}$
Or $x=5.26$
Thus the values of x are 5.26.
Thus the total amount of $\text{CaO}$is 5.26 gram and the total amount of $\text{BaO}$ is $(10-x)\text{ g=(10-5}\text{.26)g=4}\text{.76 g}$
We are interested to find out the percentage of $\text{CaO}$in the mixture.
${\scriptstyle{}^{\text{0}}/{}_{\text{0}}}\text{Cao=}\dfrac{\text{5}\text{.26}}{\text{10}}\text{ }\!\!\times\!\!\text{ 100}$
Or ${\scriptstyle{}^{\text{0}}/{}_{\text{0}}}\text{Cao}=52.6{\scriptstyle{}^{\text{0}}/{}_{\text{0}}}$
Thus the percentage of calcium oxide in the mixture of calcium oxide and barium oxide is $52.6{\scriptstyle{}^{\text{0}}/{}_{\text{0}}}$

Hence, (C) is the correct option.

Note:
The number of the equivalent of acid required to neutralize the base mixture is always equal to the number of equivalents or concentration of acid present in the solution. This is also known as the law of equivalence.