Question

A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of $C{{O}_{2}}$ at T = 298.15K and P = 1 bar. If molar volume of $C{{O}_{2}}$ is 25.0 L under such conditions, what is the percentage of sodium bicarbonate in each tablet?
[Molar mass of $NaHC{{O}_{3}}=84gmo{{l}^{-1}}$]

Answer 1

Hint: To solve this question, we should have knowledge about the stoichiometric coefficients of the reaction taking place and;
Density-
Density is the ratio of mass and volume of a substance. It is given as,
     \[Density=\dfrac{Mass}{Volume}\]
If, molar volume is given for a compound having some molar mass then density is given as,
     \[Density=\dfrac{mol.mass}{mol.volume}\]

Complete answer:
Let us know the basics to solve this problem;
Balance of chemical equation-
Balancing any chemical equation is the prime most thing that should be done when we solve the problems based on the same. This is required as we get stoichiometric coefficients when we balance the equation and hence, solving the problem becomes easier.
Now, let us move towards the illustration,
Given data-
Mass of tablet = 10 mg
Volume of $C{{O}_{2}}$ released = 0.25 ml
Molar volume of $C{{O}_{2}}$ = 25 L
Now, let us see the reaction taking place,
Reaction is between a weak acid and a mild base, so products formed will be sodium salt of acid, water and carbon dioxide.
     \[2NaHC{{O}_{3}}+{{H}_{2}}{{C}_{2}}{{O}_{4}}\to 2C{{O}_{2}}+N{{a}_{2}}{{C}_{2}}{{O}_{4}}+{{H}_{2}}O\]
Molar mass of $C{{O}_{2}}$ = 44 gm/mol
Thus, the density can be given as,
$Density=\dfrac{44gm/mol}{25L/mol}=1.76gm/L$
The mass of $C{{O}_{2}}$ released can be calculated as,
$Mass=Density\times Volume=1.76gm/L\times 0.25\times {{10}^{-3}}L$
Mass of $C{{O}_{2}}$ = $0.44\times {{10}^{-3}}$ gm
From this, moles of $C{{O}_{2}}$ produced,
Moles = $\dfrac{0.44\times {{10}^{-3}}}{44}={{10}^{-5}}$
Now, according to the stoichiometry of the reaction we can say that,
Moles of $C{{O}_{2}}$ = Moles of $NaHC{{O}_{3}}$ = ${{10}^{-5}}moles$
We have molar mass of $NaHC{{O}_{3}}$ = 84 gm/mol
Thus, mass of $NaHC{{O}_{3}}$= $84\times {{10}^{-5}}gm$= 0.84 mg
So, now percent mass of $NaHC{{O}_{3}}$ in the tablet can be given as,
$%mass=\dfrac{0.84mg}{10mg}\times 100=8.4%$
Therefore, 8.4% of $NaHC{{O}_{3}}$ is present in the tablet.

Note:
The mass percent is one of the ways of representing concentration of a component in the mixture. Do note to use proper units while solving this type of problem.