Respuesta :
Explanation:
The given data is as follows.
Mass of mixture = 0.3471 g
As the mixture contains oxalic acid and benzoic acid. So, oxalic acid will have two protons and benzoic acid has one proton.
This means oxalic acid will react with 2 moles of NaOH and benzoic acid will react with 1 mole of NaOH.
Hence, moles of NaOH in 97 ml = [tex]\frac{0.1090 \times 97}{1000}[/tex]
= [tex]10.573 \times 10^{-3} mol[/tex]
Moles of HCl in 21.00 ml = [tex]\frac{0.2060 \times 21}{1000}[/tex]
= [tex]4.326 \times 10^{-3}[/tex] mol
Therefore, total moles of NaOH that reacted are as follows.
[tex]10.573 \times 10^{-3} mol[/tex] - [tex]4.326 \times 10^{-3} mol[/tex]
= [tex]6.247 \times 10^{-3}[/tex] mol
So, total 3 mole of NaOH will react with 1 mole of mixture. Therefore, number of moles of NaOH reacted with benzoic acid is as follows.
[tex]\frac{6.247 \times 10^{-3}}{3}[/tex]
= [tex]2.082 \times 10^{-3}[/tex] mol
Since, molar mass of NaOH is 40 g/mol. Therefore, calculate the mass of NaOH as follows.
[tex]2.082 \times 10^{-3} mol \times 40 g/mol[/tex]
= [tex]83.293 \times 10^{-3}[/tex] g
= 0.0832 g
Whereas molar mass of benzoic acid is 122 g/mol.
Therefore, 40 g NaOH = 122 g benzoic acid
So, 0.0832 g NaOH = [tex]\frac{122 g}{40 g} \times 0.0832 g[/tex]
= 0.253 g
Hence, calculate the % mass of benzoic acid as follows.
[tex]\frac{0.253 g}{0.3471 g} \times 100[/tex]
= 73.10%
Thus, we can conclude that mass % of benzoic acid is 73.10%.
