Question

Butane, the fuel used in cigarette lighters, burns according to the equation: 2 C4H10 (g) + 13 O2 (g)  8 CO2 (g) + 10 H2O(g) H = – 5316 kJ a) Calculate the mass of oxygen that must react in order for this reaction to generate 2150 kJ of heat b) Calculate the amount of heat, including sign, that is transferred when 75.0 g of butane react completely.

Answer 1

2C4H10 + 13O2 ------------> 8CO2 + 10H2O delta H = -5316kJ

That is 2 moles of butane on reaction with 13 moles of oxygen completely, give 5316 kJ of heat

To generate 2150Kj of heat we need

= 2150kJ x 13 moles /5316kJ

= 5.2577 moles of oxygen

Thus mass of oxygen required to produce 2150kJ of heat =

= 5.2577 moles of oxygen x 32g/mol

= 168.25 g of oxygen

part b) According to the equation when

2 moles ( 2x58g/mol) of butane is burnt completely the heat produced = 5316kJ

Thus 75 g of butane on burning completely produces =

= 75 gx5316kJ/(2x58) g

= 3437 kJ

Thus heat produced = -3437 kJ