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Stoichiometry

Stoichiometry. -the study of the ____________ relationships between the amounts of __________ used and the amounts of __________ formed during a __________ __________. quantitative. reactants. products. chemical. reaction. I. Converting Moles of Reactant to Mass of Product

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Stoichiometry

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  1. Stoichiometry -the study of the ____________ relationships between the amounts of __________ used and the amounts of __________ formed during a __________ __________ quantitative reactants products chemical reaction I. Converting Moles of Reactant to Mass of Product and from Mass of Product to Moles of Reactant What mass, in grams, of chlorine gas can be produced from the decomposition of 2.50 moles of Sodium chloride? 2NaCl 2Na + 1Cl2 1 mole Cl2 70.906 g Cl2 2.50 moles NaCl x ___________ x ___________ = 88.6 g Cl2 2 mole NaCl 1 mole Cl2 conversion factor How many moles of 2,2,3-trimethylpentane are required to produce 100.0 grams of water in a complete combustion reaction? 2 C8H18 + 25 O2 16 CO2 + 18 H2O 1 mole H2O 2 moles C8H18 100.0 g H2O x ___________ x _____________ = 0.6166 moles C8H18 18.02 g H2O 18 mole H2O

  2. Stoichiometry I. Converting Moles of Reactant to Mass of Product and from Mass of Product to Moles of Reactant What mass, in grams, of citric acid (H3C6H5O7) can be produced from the fermentation of 7.500 moles of sucrose (C12H22O11) in air? 1 C12H22O11 + 3 O2 2 H3C6H5O7 + 3 H2O 2 mole H3C6H5O7 192.12 g H3C6H5O7 7.500 moles C12H22O11 x ______________ x _________________ = 2882 g H3C6H5O7 1 mole C12H22O11 1 mole H3C6H5O7 How many moles of Copper(II) sulfate are required to produce 50.0 grams of Zinc sulfate in a single displacement reaction? 1 CuSO4 + 1 Zn 1 ZnSO4 + 1 Cu 1 mole ZnSO4 1 mole CuSO4 50.0 g ZnSO4 x ___________ x _____________ = 0.310 moles CuSO4 161.5 g ZnSO4 1 mole ZnSO4

  3. Stoichiometry II. Converting Mass of Reactant to Mass of Product and from Mass of Product to Mass of Reactant What mass, in grams, of water can be obtained from the decomposition of 25.0 grams of Ammonium nitrate? 1NH4NO3 1N2O + 2H2O 1 mole NH4NO3 2 moles H2O 18.02 grams H2O 25.0 grams NH4NO3 x ______________ x ___________ x _____________ = 11.3 g H2O 80.04 grams NH4NO3 1 mole NH4NO3 1 mole H2O conversion factor What mass, in grams, of methane is required to produce 50.0 grams of chloroform? 1CH4 + 3Cl2 1CHCl3 + 3HCl 1 mole CHCl3 1 mole CH4 16.043 grams CH4 50.0 grams CHCl3 x ______________ x ___________ x ______________ = 6.72 g CH4 119.38 grams CHCl3 1 mole CHCl3 1 mole CH4

  4. Stoichiometry II. Converting Mass of Reactant to Mass of Product and from Mass of Product to Mass of Reactant What mass, in grams, of nitrogen can be obtained from the decomposition of 100.0 grams of Sodium azide? 2NaN3 3N2 + 2Na 1 mole NaN3 3 moles N2 28.014 grams N2 100.0 grams NaN3 x ______________ x ___________ x _____________ = 64.64 g N2 65.011 grams NaN3 2 mole NaN3 1 mole N2 What mass, in grams, of Hydrogen is required to produce 45.0 grams of methanol, in a synthesis reaction with Carbon monoxide ? 1CO + 2H2 1CH3OH 1 mole CH3OH 2 mole H2 2.016 grams H2 45.0 grams CH3OH x ______________ x ___________ x ______________ = 5.66 g H2 32.042 grams CH3OH 1 mole CH3OH 1 mole H2

  5. Stoichiometry III. Determining the Limiting Reactant If 200.0 grams of sulfur react with 100.0 grams of chlorine in a synthesis reaction, what mass, in grams, of Disulfur dichloride is produced? 1S8 + 4Cl2 4S2Cl2 1 mole S8 200.0 g S8 x ___________ = 0.7796 moles S8 _____________ = 0.7796 = In excess 256.53 g S8 1 mole S8 1 mole Cl2 100.0 g Cl2 x ___________ = 1.410 moles Cl2 _____________ = 0.3525 = Limiting reactant 70.906 g Cl2 4 moles Cl2 1 mole Cl2 4 moles S2Cl2 135.038 grams S2Cl2 100.0 grams Cl2 x ______________ x ___________ x _______________ = 190.4 g S2Cl2 70.906 grams Cl2 4 moles Cl2 1 mole S2Cl2

  6. Stoichiometry III. Determining the Limiting Reactant If 25.0 grams of phosphorus react with 50.0 grams of oxygen in a synthesis reaction, what mass, in grams, of Tetraphosphorus decoxide is produced? 1P4 + 5O2 1P4O10 1 mole P4 25.0 g P4 x ___________ = 0.202 moles P4 _____________ = 0.202 = Limiting reactant 123.90 g P4 1 mole P4 1 mole O2 50.0 g O2 x ___________ = 1.56 moles O2 _____________ = 0.312 = In excess 31.998 g O2 5 moles O2 1 mole P4 1 mole P4O10 283.89 grams P4O10 25.0 grams P4 x ______________ x ___________ x _______________ = 57.3 g P4O10 123.90 grams P4 1 mole P4 1 mole P4O10

  7. Stoichiometry IV. Calculating Percent Yield -the ________ ______ is the ______ of the _______ ______ to the ____________ ______ expressed as a ________ percent yield ratio actual yield Actual Yield theoretical yield _____________ Percent Yield = x 100 Theoretical Yield percent If 0.500 grams of Silver nitrate react with 0.500 grams of Potassium chromate and 0.455 grams of Silver chromate is produced, what is the percent yield of Silver chromate? 2AgNO3(aq) + 1K2CrO4(aq) 1Ag2CrO4(s) + 2KNO3(aq) 1 mole AgNO3 0.500 g AgNO3 x ___________ = 0.00294 moles AgNO3 _________________ = 0.00147 = Limiting reactant 169.87 g AgNO3 2 moles AgNO3 1 mole K2CrO4 0.500 g K2CrO4 x ___________ = 0.00257 moles K2CrO4 _________________ = 0.00257 = In excess 194.19 g K2CrO4 1 moles K2CrO4 1 mole AgNO3 1 mole Ag2CrO4 331.728 grams Ag2CrO4 0.500 g AgNO3 x ______________ x ___________ x __________________ = 0.488 g Ag2CrO4 169.87 grams AgNO3 2 mole AgNO3 1 mole Ag2CrO4 Theoretical Yield Actual Yield 0.455 g Ag2CrO4 _____________ _____________ Percent Yield = x 100 Percent Yield = x 100 = 93.2% Theoretical Yield 0.488 g Ag2CrO4

  8. Stoichiometry IV. Calculating Percent Yield If 40.0 grams of Hydrogen fluoride react with 40.0 grams of Silicon dioxide and 45.8 grams of Dihydrogen hexafluorosilicate is produced, what is the percent yield of Dihydrogen hexafluorosilicate? 1SiO2(s) + 6HF(aq) 1H2SiF6(aq) + 2H2O(l) 1 mole HF 40.0 g HF x ___________ = 2.00 moles HF ___________ = 0.333 = Limiting reactant 20.006 g HF 6 moles HF 1 mole SiO2 40.0 g SiO2 x ___________ = 0.666 moles SiO2 _________________ = 0.666 = In excess 60.084 g SiO2 1 moles SiO2 1 mole HF 1 mole H2SiF6 144.09 grams H2SiF6 40.0 g HF x __________ x ___________ x __________________ = 48.0 g H2SiF6 20.006 g HF 6 mole HF 1 mole H2SiF6 Theoretical Yield Actual Yield 45.8 g H2SiF6 _____________ _____________ Percent Yield = x 100 Percent Yield = x 100 = 95.4% Theoretical Yield 48.0 g H2SiF6

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