Question

Azobenzene (C6H5N)2 is produced from nitrobenzene (C6H5NO2) and triethyleneglycol (C6H14O4) in the presence of Zn and KOH according to the following balanced equation:



2 C6H5NO2 + 4 C6H14O4 → (C6H5N)2 + 4 C6H12O4 + 4 H2O


If 30.7 g of C6H5NO2 and 85.0 g of C6H14O4 are used and the reaction goes to completion, calculate the mass, in gram, of azobenzene produced by the reaction. Round your final answer to three significant figures.

Answer 1

$n(C6H14O4)=85(g)/150.17(g/mol)=0.56 moles$

$n(C6H5NO2)=30.7(g)/123(g/mol)=0.25 moles$

Nitrobenzene is a limiting reagent.

$n((C6H5N)2)=n(C6H5NO2)/2=0.25moles/2=0.125moles$

$m(C6H5N)2)=182.2(g/mol)*0.125moles=22.7g$

Answer: 22.7g