# u8 - Unit 8 Stoichiometry involves finding amts of...

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Unit 8: Stoichiometry -- involves finding amts. of reactants & products in a reaction

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amount of RA and/or RB you must use amount of P1 or P2 you need to produce amount of P1 or P2 that will be produced amount of RA or RB amount of RB (or RA) that is needed to react with it amount of RA (or RB) …one can find the… Given the… What can we do with stoichiometry? For generic equation: RA + RB P1 + P2
4 patties + ? Governing Equation: 2 patties + 3 bread 1 Big Mac® excess + 18 bread ? ? + ? 25 Big Macs® 6 bread 75 bread 50 patties 6 Big Macs®

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Use coefficients from balanced equation MOLE (mol) Mass (g) Particle (at. or m’c) 1 mol = molar mass (in g) Volume (L or dm3) 1 mol = 22.4 L 1 mol = 22.4 dm3 1 mol = 6.02 x 1023 particles SUBSTANCE “A” Stoichiometry Island Diagram MOLE (mol) Mass (g) 1 mol = molar mass (in g) Volume (L or dm3) 1 mol = 22.4 L 1 mol = 22.4 dm3 1 mol = 6.02 x 1023 particles SUBSTANCE “B” Particle (at. or m’c)
2 __TiO2 + __Cl2 + __C __TiCl4 + __CO2 + __CO How many mol chlorine will react with 4.55 mol carbon? 3 mol C 4 mol Cl2 4.55 mol C = 6.07 mol Cl2 What mass titanium (IV) oxide will react with 4.55 mol carbon? ( ) = 242 g TiO2 4 3 2 2 1 ( ) C Cl 2 C TiO2 3 mol C 2 mol TiO2 4.55 mol C ( ) 1 mol TiO2 79.9 g TiO2

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3. The units on the islands at each end of the bridge being crossed appear in the conversion factor for that bridge. How many molecules titanium (IV) chloride can be made from 115 g titanium (IV) oxide? TiCl 4 TiO2 ( ) = 8.7 x 1023 m’c TiCl4 115 g TiO2 1 mol TiO2 79.9 g TiO2 ( ) 2 mol TiO2 2 mol TiCl4 ( ) 1 mol TiCl4 6.02 x 1023 m’c TiCl4 Island Diagram helpful reminders: 2. The middle bridge conversion factor is the only one that has two different substances in it. The conversion factors for the other six bridges have the same substance in both the numerator and denominator. 1. Use coefficients from the equation only when crossing the middle bridge . The other six bridges always have “1 mol” before a substance’s formula. 1 m o l 1 mol 1 mol coeff.
2 Ir + Ni3P2 3 Ni + 2 IrP If 5.33 x 1028 m’cules nickel (II) phosphide react w/excess iridium, what mass iridium (III) phosphide is produced? Ni3P2 IrP ( ) = 3.95 x 107 g IrP 1 mol IrP 223.2 g IrP ( ) 1 mol Ni3P2 2 mol IrP ( ) 1 mol Ni3P2 6.02 x 1023 m’c Ni3P2 5.33 x 1028 m’c Ni3P2 How many grams iridium will react with 465 grams nickel (II) phosphide? ( ) = 751 g Ir 1 mol Ir 192.2 g Ir ( ) 1 mol Ni3P2 2 mol Ir ( ) 1 mol Ni3P2 238.1 g Ni3P2 465 g Ni3P2 Ni3P2 Ir

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2 mol Ir How many moles of nickel are produced if 8.7 x 1025 atoms of iridium are consumed?
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## This note was uploaded on 11/09/2011 for the course CHEM 110 taught by Professor Sullivan during the Fall '10 term at BYU.

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u8 - Unit 8 Stoichiometry involves finding amts of...

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