What mass of butane, C4H10, would need to be combusted in order to convert 100.0 g of ice that is initially at 0.0°C to steam at 175.0°C? Assume that all of the heat generated goes directly into heating the ice. Enthalpy of combustion of gaseous butane is −2874 kJ/mol. Heat of vaporization of water is 2260 J/g, and heat of fusion of ice is 333 J/g. specific heat(c) of water = 4.18 J/g°C.

specific heat(c) of steam = 2.05 J/g°C.

Provide work.

the mass is m (g) , then we have m/58*2874 x10^3 = 100*(333+4.18*100+2260+2.05*75)  = 316475

m/58 = 316475/(2874×10^3) =  0.1101m = 6.387 g 

In respects to the question and the work you provided, what is the equation used to derive the amount of grams of butane? Thanks.