How can I balance this C3H8 + O2 --------> CO2 + H2O + heat?
- Dr^valentineLv 41 decade agoFavorite Answer
It is the heat of combustion of C3H8.
So, you must not balance the C3H8 (1 mole of C3H8)
C3H8 + O2 --------> 3CO2 + H2O + heat
C3H8 + O2 --------> 3CO2 +4 H2O + heat
C3H8 + 5O2 --------> 3CO2 +4 H2O + heat
- 1 decade ago
Try multiples of 3 on the CO2..You then need 8 H's on both sides so try a 4 on the H2O.. You then need an extra 5 O's on the left-hand side
C3H8 + 5 O2 => 3 CO2 + 4 H2O + heat
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- PatriciaLv 44 years ago
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1A) balance the equation 1 C3H8(l) + 5 O2(g) -> 3 CO2(g) + 4 H2O(l) 2A) look up the heat of formation values C3H8(l) = −119.8 kJ/mol O2 (g) = 0 kJ/mol CO2(g) = -393.5 kJ/mol H2O(l) = -285.8 kJ/mol 3A) use the balanced equation and the heat of formation values to calculate Hrxn -Hrxn = products - reactants -multiply each number by the number of moles in the balanced equation [4*(-285.8 kJ/mol)+3*(-393.5 kJ/mol)] - [1*(−119.8 kJ/mol)] = -2203.9 kJ/mol B: set up a stoichiometric equation 1B) in setting up equation, change g. into moles (MM=44.0962 mol/g) 715 g * ( 1 mol/ 44.0962g) = 16.21455 mol C3H8 2B) using # of moles, set up equation with the calculated Hrxn 16.21455 mol C3H8 * (-2203.9 kJ/mol) = -35735 kJ 3B) with sig figs, it would be -35700
- 1 decade ago
Oohh.. the dreaded hydrocarbon. This guy ^^^^^ has the correct answer.