If #"1500 g"# of water was heated by the combustion of #"1 g"# of sucrose to go from #25.0^@ "C"# to #28.3^@ "C"#, what is the enthalpy of combustion in #"kJ/mol"#?

1 Answer

Answer:

-7080 kJ /mol of sucrose

Explanation:

Heat released

#"Increase of water temperature = 3.3 K = (28.3-25.0) K"#

#"1500 g water" = "1500 g"/"18.02 g/mol" = "83.24 mol"# of water

#q="83.24 mol" * "75.4 J·mol"^"-1""K"^"-1" *3.3 "K" = "20700 J"#

The amount of heat released by burning 1 g of sucrose is 20.7 kJ per gram of sucrose.

However, 1 mol of sucrose is 342 g.

Therefore #Delta H# can be computed:

#DeltaH = "-20.7 kJ/g" * "342 g/mol" = "-7080 kJ/mol"#