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:

2 C4H10   +      13 O2       →      8 CO2   +     10H2O
Given, Heat of formation of C4H10 =  = -125.7 kJ/molHeat of formation of water =  = -241.82 kJ/molHeat 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 question is incomplete. complete question is attached belowanswer: correct answer is option b  i.e.  solid particles  reason: filter paper has small pores, through which smaller particles can pass through. however, solid particles  have relatively larger size. hence, they are least likely to pass through.