Chap 3 Sec 4

# Chap 3 Sec 4 - Chap 3 Sect 4 sm Page 1 10/08 3.4...

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Unformatted text preview: Chap 3 Sect 4 sm.doc Page 1 10/08 3.4 Calculating Amounts of Reactant and Product Mole to Mole Ratios Use the balanced equation: 2 C 3 H 8 O + 9 O 2-----> 6 CO 2 + 8 H 2 O If we have 3.25 moles of C 3 H 8 O how many moles of oxygen will be used? 2 moles C 3 H 8 O = 9 moles O 2 Treat like a conversion factor!!! 3.26 mole C 3 H 8 O x 9 moles O 2 2 mole C 3 H 8 O = 14.67 moles O 2 = 14.7 moles O 2 or 3.26 mole C 3 H 8 O x 9 O 2 2 C 3 H 8 O = 14.7 moles O 2 Any pair can be used: We can go from CO 2 formed to C 3 H 8 O We can go from CO 2 formed to H 2 O Mass to mass method mass -----> moles -----> moles ------> mass Example 1: If 2.85 g of NH 3 are burned, how much H 2 O will be formed? Given: 4 NH 3 + 5 O 2-----> 4 NO + 6 H 2 O (MW) 17.04 18.02 Method: a) g NH 3-----> moles NH 3 b) moles NH 3-----> moles H 2 O c) moles H 2 O -----> g H 2 O--------------------------------------------------------------- a) 2.85 g NH 3 x 1 mole 17.04g = 0.16725.. mol NH 3 Chap 3 Sect 4 sm.doc Page 2 10/08 b)...
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Chap 3 Sec 4 - Chap 3 Sect 4 sm Page 1 10/08 3.4...

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