Stoichiometry (68)

# Stoichiometry (68) - Chapter 10 Mol Factors Calculations...

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1 Chapter 10 Mol Factors Calculations with Equations Limiting Reactions Percent Yield

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2 Mole-Mole Factor Shows the mole-to-mole ratio between two of the substances in a balanced equation Derived from the coefficients of any two substances in the equation
3 Writing Mole Factors 4 Fe + 3 O 2 2 Fe 2 O 3 Fe and O 2 4 mol Fe and 3 mol O 2 3 mol O 2 4 mol Fe Fe and Fe 2 O 3 4 mol Fe and 2 mol Fe 2 O 3 2 mol Fe 2 O 3 4 mol Fe

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4 O 2 and Fe 2 O 3 3 mol O 2 and 2 mol Fe 2 O 3 2 mol Fe 2 O 3 3 mol O 2
5 Learning Check S1 3 H 2 (g) + N 2 (g) 2 NH 3 (g) A. A mol factor for H 2 and N 2 is 1) 3 mol N 2 2) 1 mol N 2 3) 1 mol N 2 1 mol H 2 3 mol H 2 2 mol H 2 B. A mol factor for NH 3 and H 2 is 1) 1 mol H 2 2) 2 mol NH 3 3) 3 mol N 2

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6 Solution S1 3 H 2 (g) + N 2 (g) 2 NH 3 (g) A. A mol factor for H 2 and N 2 is 2) 1 mol N 2 3 mol H 2 B. A mol factor for NH 3 and H 2 is 2) 2 mol NH 3
7 Chemical Calculations 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe 2 O 3 are produced when 6.0 moles O 2 react? 6.0 mol O 2 x mol Fe 2 O 3 = 4.0 mol Fe 2 O 3 mol O 2

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8 Learning Check 2 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe are needed to react with 12.0 mol of O 2 ? 1) 3.00 mol Fe 2) 9.00 mol Fe 3) 16.0 mol Fe
9 Solution S2 4 Fe + 3 O 2 2 Fe 2 O 3 12.0 mol O 2 x mol Fe = 16.0 mol Fe mol O 2 4 3

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## This note was uploaded on 11/09/2011 for the course CHEM 100 taught by Professor Sdfsdf during the Winter '08 term at BYU.

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Stoichiometry (68) - Chapter 10 Mol Factors Calculations...

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