Question

) If you started with 2.3182 g of 3-nitrophthalic acid and had a 93% yield of 3-nitrophthalhydrazide, how many grams of 3-nitrophthalhydrazide were recovered

Answer 1

Answer: 1.8124 g of 3-nitrophthalhydrazide were recovered.

Explanation:

The balanced chemical reaction will be :

$C_8H_5NO_6+N_2H_4\rightarrow C_8H_5N_3O_4+2H_2O$

moles of 3-nitrophthalic acid  = $\frac{\text {given mass}}{\text {molar mass}}=\frac{2.3182g}{211.13g/mol}=0.0110mol$

As 1 mole of 3-nitrophthalic acid gives = 1 mole of 3-nitrophthalhydrazide

0.0110 moles of 3-nitrophthalic acid gives = $\frac{1}{1}\times 0.0110=0.0110$ mole of 3-nitrophthalhydrazide

mass of 3-nitrophthalhydrazide = $moles\times {\text {molar mass}}=0.0110mol\times 177.16g/mol=1.9488g$

As the percentage yield is 93% , the mass of 3-nitrophthalhydrazide recovered = $\frac{1.9488\times 93}{100}=1.8124g$

Therefore 1.8124 g of 3-nitrophthalhydrazide were recovered.