Respuesta :
Answer: 1.8124 g of 3-nitrophthalhydrazide were recovered.
Explanation:
The balanced chemical reaction will be :
[tex]C_8H_5NO_6+N_2H_4\rightarrow C_8H_5N_3O_4+2H_2O[/tex]
moles of 3-nitrophthalic acid = [tex]\frac{\text {given mass}}{\text {molar mass}}=\frac{2.3182g}{211.13g/mol}=0.0110mol[/tex]
As 1 mole of 3-nitrophthalic acid gives = 1 mole of 3-nitrophthalhydrazide
0.0110 moles of 3-nitrophthalic acid gives = [tex]\frac{1}{1}\times 0.0110=0.0110[/tex] mole of 3-nitrophthalhydrazide
mass of 3-nitrophthalhydrazide = [tex]moles\times {\text {molar mass}}=0.0110mol\times 177.16g/mol=1.9488g[/tex]
As the percentage yield is 93% , the mass of 3-nitrophthalhydrazide recovered = [tex]\frac{1.9488\times 93}{100}=1.8124g[/tex]
Therefore 1.8124 g of 3-nitrophthalhydrazide were recovered.
