C6H6+15O2-12CO2+6H2O+ 1562kcal if you want to obtain 125kcal of heat by reacting benzene with oxygen, how many grams of benzene must you use?

1 Answer
Truong-Son N.
Feb 28, 2018

I got #"6.25 g benzene"#.

For the reaction AS-WRITTEN, we have:

#"C"_6"H"_6(s) + 15"O"_2(g) -> 12"CO"_2(g) + 6"H"_2"O"(g)#, #DeltaH_(rxn)^@ = -"1562 kcal"#

That means the reaction releases #bb("1562 kcal"ul"/mol benzene")#.

Energy is intensive when in units of just #"J"# or #"kJ"#... Therefore, releasing #bb"125 kcal"# would be due to combustion of:

#"1 mol benzene"/(1562 cancel"kcal") xx 125 cancel"kcal" = "0.0800 mols benzene"#

or

#0.0800 cancel("mols C"_6"H"_6) xx ("78.1134 g")/cancel("1 mol C"_6"H"_6) = color(blue)"6.25 g benzene"#

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