Concerning the reaction #C + O_2 -> CO_2#, if the #ΔH° = -393# #kJ#/#mol#, how many grams of #C# must be burned to release #275# #kJ# of heat?
1 Answer
Explanation:
Notice that the standard enthalpy change of reaction,
#"C"_text((s]) + "O"_text(2(g]) -> "CO"_text(2(g])#
More specifically, when one mole of carbon reacts with one mole of oxygen gas, one mole of carbon dioxide is produced and
Remember, a negative enthalpy change of reaction tells you that heat is being given off, i.e. the reaction is exothermic.
Now, in order to determine how much carbon is needed in order for the reaction to give off
#275 color(red)(cancel(color(black)("kJ"))) * "1 mole C"/(393 color(red)(cancel(color(black)("kJ")))) = "0.6997 moles C"#
To see how many grams of carbon would contain this many moles, use carbon's molar mass
#0.6997 color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = color(green)("8.40 g")#
So, when
#DeltaH^@ = -"275 kJ"#