Question

A mixture containing 248.1g of H2O and 32g of N2H4 reacts as follows:

7H2O2 + N2H4...........2HNO3 + 8H20

calculate the amount of HNO3 produced. If the acyual yield of HNO3 is 64g, calculate the percent yield and percent error

Answer 1

From the balance chemical equation

7 moles of H2O2 will react with 1 mole of N2H4 will give 2 moles of HNO3

No. of moles of H2O2 = 248.1 / 34.014 = 7.294 moles

No. of moles of N2H4 = 32 / 32.046 = 0.9985 moles

for each mole of N2H4, no. of moles of H2O2 needed = 7 x 0.9985 = 6.99 mole

Here H2O2 present in excess, so limiting reagent is N2H4

0.9985 moles of N2H4 will react with excess H2O2 will give product = 2 x 0.9985 = 1.997 moles of HNO3

Weight of HNO3 formed = 1.997 x 63.012 = 125.83 grams (theoretical yield)

Percentage yield = (64/125.83) x 100 = 50.86 %