Posted by desdemona on Saturday, November 26, 2011 at 2:11pm.
C3H8 + 5O2 ==> 3CO2 + 4H2O
This is a limiting reagent and we know that because amount for BOTH reactants are given. We must first determine the limiting reagent (either propane or oxygen).
3 moles C3H8 will produce how many moles CO2 (ignoring oxygen)? (3 moles C3H8 x 3 moles CO2/1 mole C3H8) = 3 x (3/1) = 9 moles CO2.
10 moles O2 will produce how much CO2 (ignoring propane). 10 moles O2 x 3 moles CO2/5 moles O2) = 10 x (3/5) = 6 moles CO2.
Obviously, both answers can't be right; the correct answer in limiting reagent problems is ALWAYS the smaller value (in this case 6 moles CO2) and the reagent producing that value is the limiting reagent (in this case the 10 moles O2). Now the problem is hos much heat is produced by burning C3H8 with 10 moles O2. The easier way is to determine how much C3H8 reacted. That will be
10 moles O2 x (1 mole C3H8/5 moles O2) = 10 x (1/5) = 2 moles C3H8.
Therefore, it C3H8 produces 2012 kJ/1 mol then it must produce 2 x that for 2 moles.
2012 kJ/mol x 2 mol = ?
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