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A $10{\text{ mg}}$ effervescent tablet containing sodium bicarbonate and oxalic acid releases $0.25{\text{ ml}}$ of ${\text{C}}{{\text{O}}_2}$ at ${\text{T = }}298.15{\text{ K}}$ and ${\text{P = }}1{\text{ bar}}$. If molar volume of ${\text{C}}{{\text{O}}_{\text{2}}}$ is $25.9{\text{L}}$ under such condition, what is the percentage of sodium bicarbonate in each tablet?
$\left[ {{\text{Molar mass of NaHC}}{{\text{O}}_3} = {\text{ }}84{\text{ g mo}}{{\text{l}}^{ - 1}}} \right]$
A.$16.8$
B.$8.4$
C.$0.84$
D.$33.6$

seo-qna
Last updated date: 17th Apr 2024
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Answer
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Hint:To solve this question, it is required to have knowledge about stoichiometric coefficients of the reaction occurring. Firstly, we are given the molar volume of carbon dioxide, from there dividing it with molar mass we shall get the density of carbon dioxide. Then, we shall find the mass of the given volume of carbon dioxide and convert it into moles using mole concept formula (given). Using the stoichiometric coefficients of the reaction we shall calculate the moles of sodium bicarbonate present and then convert it to mass. We can then calculate the percentage mass from the mass of the tablet given.

Formula used: ${\text{Moles = }}\dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$ and ${\text{Density = }}\dfrac{{{\text{mass}}}}{{{\text{volume}}}}$

Complete step by step answer:
As the reaction is occurring between a mild base and a weak acid so, the products formed will be a sodium salt of the acid, water and carbon dioxide. First, we shall write a balanced equation for the reaction occurring:
${\text{2NaHC}}{{\text{O}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}} \to 2{\text{C}}{{\text{O}}_{\text{2}}}{\text{ + N}}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
We know that $1{\text{ mole C}}{{\text{O}}_2} = 44{\text{ g mo}}{{\text{l}}^{ - 1}}{\text{ = 25}}{\text{.9L}}$
So, the density of carbon dioxide will be: ${\text{density = }}\dfrac{{{\text{44}}}}{{{\text{25}}{\text{.9}}}}{\text{ = 1}}{\text{.70 g }}{{\text{L}}^{{\text{ - 1}}}}$
From here, we can calculate the mass of carbon dioxide produced in the reaction. Thus:
${\text{mass of C}}{{\text{O}}_2} = \dfrac{{1.70 \times 0.25}}{{1000}}g$
$ \Rightarrow {\text{mass of C}}{{\text{O}}_2} = 0.425 \times {10^{ - 3}}{\text{ g}}$
 Now, we shall find the number of moles of carbon dioxide produced in this reaction.
$ \Rightarrow {\text{moles of C}}{{\text{O}}_2} = \dfrac{{0.425}}{{44}} \times {10^{ - 3}}$
$ \Rightarrow {\text{moles of C}}{{\text{O}}_2} = 0.96 \times {10^{ - 5}} \sim {10^{ - 5}}$
So, according to the reaction equation, the moles of carbon dioxide and sodium bicarbonate in the reaction are equal. So, moles of ${\text{NaHC}}{{\text{O}}_3} = {10^{ - 5}}$. As we know the molar mass of sodium bicarbonate, we shall convert the moles into mass using the mole concept formula.
$ \Rightarrow {\text{mass = 84}} \times {\text{1}}{{\text{0}}^{ - 5}}{\text{ g}}$. This can also be written as ${\text{mass = 0}}{\text{.84 mg}}$.
Now, we know the mass of sodium bicarbonate formed and also the total mass of the tablet. So, now we can find the percentage mass of sodium bicarbonate in the tablet.
${\text{\% mass = }}\dfrac{{{\text{mass of NaHC}}{{\text{O}}_3}}}{{{\text{mass of tablet}}}} \times 100$
Putting the appropriate values in the equation, we get:
${\text{\% mass = }}\dfrac{{0.84{\text{ mg}}}}{{{\text{10 mg}}}} \times 100$
${\text{\% mass = }}8.4\% $
$\therefore $The correct option is option B, i.e. $8.4\% $.

Note:In this type of question, we should be aware so as to not get confused in the units. If the mass is given in milligrams, we shall try to convert the mass as much as we can into milligrams or convert the milligrams into grams. In the reaction equation, oxalic acid is a carboxylic acid, so after reaction with a base, the hydrogen ions of the acid group shall be replaced by the sodium ion of the base.