Respuesta :
the balanced equation for the reaction between NaOH and aspirin is as follows;
NaOH + C₉H₈O₄ --> C₉H₇O₄Na + H₂O
stoichiometry of NaOH to C₉H₈O₄ is 1:1
The number of NaOH moles reacted - 0.1002 M / 1000 mL/L x 10.00 mL
Number of NaOH moles - 0.001002 mol
Therefore number of moles of aspirin - 0.001002 mol
Mass of aspirin reacted - 0.001002 mol x 180.2 g/mol = 0.18 g
However the mass of the aspirin sample is 0.132 g but 0.18 g of aspirin has reacted, therefore this question is not correct.
NaOH + C₉H₈O₄ --> C₉H₇O₄Na + H₂O
stoichiometry of NaOH to C₉H₈O₄ is 1:1
The number of NaOH moles reacted - 0.1002 M / 1000 mL/L x 10.00 mL
Number of NaOH moles - 0.001002 mol
Therefore number of moles of aspirin - 0.001002 mol
Mass of aspirin reacted - 0.001002 mol x 180.2 g/mol = 0.18 g
However the mass of the aspirin sample is 0.132 g but 0.18 g of aspirin has reacted, therefore this question is not correct.
Answer:
73.10% the percent purity of aspirin.
Explanation:
[tex]NaOH+HAsp\rightarrow NaAsp+H_2O[/tex]
Moles of NaOH:
[tex]0.1002 M=\frac{\text{moles of NaOH}}{0.010 L}[/tex]
Moles of NaOH = 0.001002 moles
According to reaction. 1 NaOH reacts with 1 mole of aspirin .
Then 0.001002 moles NaOH will reacts with :
[tex]\frac{1}{1}\times 0.001002 moles=0.001002 moles[/tex]
Mass of 0.001002 moles of aspirin:
= 0.001002 moles × 180.2 g/mol = 0.18056 g
Percentage of purity of aspirin:
[tex]\%=\frac{0.132 g}{0.18056 g}\times 100=73.10\%[/tex]
73.10% the percent purity of aspirin.
