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Scan_Doc0011 - 104 Chapter Three Stoichiometry Next we must...

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--------- 104 Chapter Three Stoichiometry Next we must take into account the fact that each mole of propane reacts with 5moles of oxygen. The best way to do this is to use the balanced equation to construct a mole ratio. In this case we want to convert from moles of propane to moles of oxygen. From the balanced equation we see that 5 moles of O 2 is required for each mole of C 3 H s , so the appropriate ratio is 5 mol O 2 I mol C3HS Multiplying the number of moles of C3HS by this factor gives the number of moles of O 2 required: 5mol O 2 = lO.9 mol O 2 2.18~X1~ Notice that the mole ratio is set up so that the moles of C 3 Hs cancel out, and the units that result are moles of O 2 . Since the original question asked for the mass of oxygen needed to react with 96.1 grams of propane, the lO.9 moles of O 2 must be converted to grams. Since the molar mass of O 2 is 32.0 g/mol, 32.0 g O 2 = 349 g O 2 lO.9 meI-02 X 1 meI-02 Therefore, 349 grams of oxygen is required to burn 96.1 grams of propane.
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