This is what I did:
moles of H2O = 3.88
delta H rxn = -2044.02
(3.88 moles)(6.01 kJ/mol) = 23.359 kJ
(3.88 moles)(75.291 J/mol/C) = 29.263 kJ
23.359kJ + 29.2636 kJ = 29.286 kJ
(44 g of C3H8)(29.286kJ/-2044.02kJ) = 0.6304 g
But I think I did something wrong??
How many grams of propane must be burned to provide the heat needed to melt a 70.0-g piece of ice...?
How many grams of propane must be burned to provide the heat needed to melt a 70.0-g piece of ice (at its melting point) and bring the resulting water to a boil (with no vapourization)? Assume a constant pressure of 1 bar and use the following data:
ΔHfo(H2O(g))=-241.83 kJ/mol, ΔHfo(O2(g))=0 kJ/mol, ΔHfo(CO2(g))=-393.5 kJ/mol, ΔHfo(C3H8(g))=-103.8 kJ/mol
Tfus(H2O)=0 oC, Tvap(H2O)=100 oC
when I tried doing it I got 0.63 but I don't think that's correct
1 answer