Question

What is the yield of 2,4-dinitrophenylhydrazine obtained from the reaction of hydrazine and 2,4-dinitrochlorobenzene if 5 grams of hydrazine and 7 grams of 2,4-dinitrochlorobenzene are used and the resulting product weighs 4.5 grams?

Answer 1

To calculate the yield of 2,4-dinitrophenylhydrazine, we first need to determine the limiting reactant and the theoretical yield.The balanced chemical equation for the reaction is:N2H4 + C6H3Cl NO2 2  C6H3 NO2 2N2H3 + HClMolar mass of hydrazine  N2H4  =  2  14.01  +  4  1.01  = 32.05 g/molMolar mass of 2,4-dinitrochlorobenzene  C6H3Cl NO2 2  = 12.01  6 + 1.01  3 + 35.45 + 14.01  2 + 16.00  4 = 202.56 g/molMolar mass of 2,4-dinitrophenylhydrazine  C6H3 NO2 2N2H3  = 12.01  6 + 1.01  3 + 14.01  2 + 1.01  3 + 16.00  4 = 198.14 g/molMoles of hydrazine = 5 g / 32.05 g/mol = 0.156 molesMoles of 2,4-dinitrochlorobenzene = 7 g / 202.56 g/mol = 0.0345 molesThe stoichiometry of the reaction is 1:1, so we can determine the limiting reactant by comparing the mole ratios:0.156 moles N2H4 / 1 = 0.1560.0345 moles C6H3Cl NO2 2 / 1 = 0.0345Since 0.0345 is smaller than 0.156, 2,4-dinitrochlorobenzene is the limiting reactant.Now we can calculate the theoretical yield:0.0345 moles C6H3Cl NO2 2   1 mol C6H3 NO2 2N2H3 / 1 mol C6H3Cl NO2 2   198.14 g/mol = 6.83 gFinally, we can calculate the percentage yield: Actual yield / Theoretical yield   100 =  4.5 g / 6.83 g   100 = 65.89%The yield of 2,4-dinitrophenylhydrazine is 65.89%.