The more butane is burnt, the more heat is produced. So calculating what quantites of butane undergo reaction in each case will show what amounts of heat are produced.
1) n(C4H10) = 1 mol
2) n(C4H10) = "{\\frac {2} {13}}" mol accroding to equation, as n(O2) = 1 mol and n(C4H10) = "{\\frac {2} {13}}"*n(O2)
3) n(C4H10) = "{\\frac {1} {4}}" mol according to equation, as n(CO2 produced) = 1 mol and n(C4H10) = "{\\frac {2} {8}}"*n(CO2)
4) n(C4H10) = "{\\frac {1} {5}}" mol according to equation, as n(H2O produced) = 1 mol and n(C4H10) = "{\\frac {2} {10}}"*n(H2O)
the least amount of butane involved is in the second case, so the least amount of heat generated in 2 situation
Comments
Leave a comment