Posted by od on Saturday, April 7, 2012 at 2:41pm.
You may use a shortcut to do these stoichiometry problems when only gases are involved.
C3H8 + 5O2 ==> 3CO2 + 4H2O
I do these in two parts.
1. How much CO2 is produced if I have 10L C3H8 and all the oxygen I need? That is
10L propane x (3 moles CO2/1 L C3H8) = 10 x 3/1 = 30 L CO2
2. How much CO2 is produced if I have 10 L O2 and all the propane I need? That is
10 L O2 x (3 mols CO2/5 mols O2) = 10 x 3/5 = 6 L CO2.
Both answers can't be correct; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent. Therefore, O2 is the limiting reagent (and we can produce 6L CO2 although that's isn't the question). Now that we know the limiting reagent, we can calculate how much of the other one is used.
10 L O2 x (1 mol C3H8/5 mol O2) = 10 x 1/5 = 2 L propane used.
We started with 10, used 2, we must have 8 left over. Book is right.
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