Question

50.0 grams of C_16H_34 are burned at 1 bar pressure with insufficient oxygen for complete combustion to CO_2(g). As a result, some of the reactant forms CO_(g) instead via the reaction: C_16H_34(s) + O_2(g) Right arrow CO_(g) + H_2O(I) (unbalanced) And some of the reactant forms CO_2(g). The heat generated in the process was found to be-1945.2 kJ. Compute the fraction of hexadecane that formed CO(g). Compute the internal energy change for this system.
Both a and b

Answer 1

a) balanced equation :

2C16H34 + 47O2 ---> 28CO2 + 4CO + 34H2O

dHrxn = -1945.2 kJ

moles of C16H34 = 50 g/226.44 g/mol = 0.22 mols

2 mole of C16H34 reacts to form 4 moles of CO

So fraction of hexadecane forming CO = 0.22 x 4/2 = 0.44 mols

b) Change in internal energy = -1945.2/0.44 = 4420.91 kJ/mol