Butane c4 h10 (-1.jpghf = –125.7), combusts in the presence of oxygen to form co2 (g) (mc016-2.jpghf = –393.5 kj/mol), and h2 o(g) (mc016-3.jpghf = –241.82) in the reaction: mc016-4.jpg what is the enthalpy of combustion, per mole, of butane?

Following is the balanced reaction for combustion of butane,

2 C4H10   +      13 O2       →      8 CO2   +     10H2O

Given, Heat of formation of C4H10 =  = -125.7 kJ/mol
Heat of formation of water =  = -241.82 kJ/mol
Heat of formation of CO2 =  = -393.5 kJ/mol

Thus, enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants
                                               = 8 + 10 - 2 
                                               = (-393.5X8) + (-241.82X10) - (-125.7X2)
                                               = -5314.8 kJ/mol

This energy would result in combustion of 2 moles of butane
∴ For 1 mole, energy required for combustion of butane = -5314.8/2 = -2657.4 kJ/mol

The correct answer is c. orbital shape. "the angular momentum quantum number l describes the shape of an orbital. it corresponds to each orbital type, i.e. (0,1,2,−1)=(s,p,d,f,g,"