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Stoichiometry

Stoichiometry. Mole Relationships, Mass Relationship, and Percent Yield. Mole Relationships. coefficients can be put in the form of MOLE RATIOS. H 2 + N 2  NH 3. This number of moles of Hydrogen. But what does it mean?. UN. BALANCED. 2. 2. 3. 3. 1.

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Stoichiometry

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  1. Stoichiometry Mole Relationships, Mass Relationship, and Percent Yield

  2. Mole Relationships coefficients can be put in the form of MOLE RATIOS H2 + N2  NH3 This number of moles of Hydrogen ... But what does it mean? UN BALANCED 2 2 3 3 1 ... reacts with this number of moles of nitrogen ... ... to yield this number of moles of ammonia ... 3 mol H2 1 mol N2 = 3 mol H2 2 mol NH3 = 1 mol N2 2 mol NH3

  3. Mole Ratio Example 4 Fe(s) + 3 O2(g)  2 Fe2O3 (s) What are all the possible mole ratios? 4 mol Fe 3 mol O2 4 mol Fe 2 mol Fe2O3 3 mol O2 2 mol Fe2O3 or or How many moles of iron(III) oxide are formed by the complete oxidation of 6.2 mol of iron? 6.2 mol Fe x 2 mol Fe2O3 4 mol Fe = 3.1 mol Fe2O3

  4. Mass Relationship Mole  Mass and Mass  Mole need to find this know this Moles of A  Grams of A carried out using molar mass as a conversion factor

  5. Mass Relationship Mole  Mole conversion need to find this know this Moles of A  Moles of B use mole ratio as a conversion factor

  6. Mass Relationship Mass A  Mass B conversions frequently needed can’t be carried out directly 1. Convert mass A into moles A Use mole ratio to find moles of B Convert moles of B into mass B

  7. Mass Relationship Mass A  Mass B conversions Overall Steps: Write balanced chemical equation Choose molar masses and mole ratios to convert the known information into the needed information Set up factor label expression and calculate

  8. Examples CaCl2(aq) + Na2C2O4(aq) CaC2O4(s) + 2NaCl(aq) The previous reaction produced 0.222 g of calcium oxate (CaC2O4). What mass of calcium chloride was used as a reactant? 0.222 g CaC2O4 x 1 mol CaCl2 1 molCaC2O4 x 110.98 g CaCl2= 1 mol CaCl2 x 1 mol CaC2O4 128.10 g CaC2O4 x molar mass x molar ratio x molar mass 0. 192 g CaCl2

  9. Examples 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) How many grams of HNO3 are produced for every 2.50 mol of NO2 that reacts? x 2 mol HNO3 3 mol NO2 x 63.01 g HNO3= 1 mol HNO3 2.50 mol NO2 105.02 g HNO3 x molar ratio x molar mass

  10. Examples Hydrogen flouride is one of the few substances that react with glass (silicone dioxide, SiO2) HF(g) + SiO2(s)  SiF4(g) + H2O(l) • How many moles of HF will react completely with 9.90 g of SiO2? 4 2 9.90 g SiO2 x 4 mol HF 1 mol SiO2 x 1 mol SiO2 60.08 g SiO2 = 0.659 mol HF x molar mass x molar ratio

  11. Examples Hydrogen flouride is one of the few substances that react with glass (silicone dioxide, SiO2) HF(g) + SiO2(s)  SiF4(g) + H2O(l) • How many grams of water will be produced by the reaction of 23.0 g of SiO2? 4 2 23.0 g SiO2 x 2 mol H2O 1 mol SiO2 x 18.02 g H2O= 1 mol H2O x 1 molSiO2. 60.08 g SiO2 x molar mass x molar ratio x molar mass 13.8 g H20

  12. Percent Yield If in the previous reaction of glass and hydrogen flouride, 5.44 g of water was produced instead, what was the percent yield? Actual Yield Theoretical Yield x 100 % = Percent yield (calculated) 5.44 g H2O x 100 % = 39.4 % 13.8 g H2O

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