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 condition, what is the percentage of sodium bicarbonate in each tablet? [Molar mass of $NaHC{O}_3=84gmo{l}^{-1}$ ]

• A. 0.84
• B. 8.4
• C. 16.8
• D. 33.6

Answer 1

The correct option is B 8.4

$\underset{2mol}{2NaHC{O}_3}+\underset{1mol}{H_2{C}_2{O}_4}\to \underset{2mol}{2C{O}_2}+N{a}_2{C}_2{O}_4+H_{2}O$
In the reaction, number of moles of $C{O}_2$ produced,
$n=\frac{pV}{RT}=\frac{1bar\times 0.25\times {10}^{-3}L}{0.082Latm{K}^{-1}mo{l}^{-1}\times 298.15K}$
$=1.02\times {10}^{-5}mol$
Number of moles of $NaHC{O}_3=\frac{\text{Weight of}NaHC{O}_3}{\text{Molecular mass of}NaHC{O}_3}$
$\therefore w_{NaHC{O}_3}=1.02\times {10}^{-5}\times 84\times {10}^{3}mg$
$=0.856mg$
$\Rightarrow NaHC{O}_3\%=\frac{0.856}{10}\times 100=8.56\%$