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Chemical Stoichiometry

Chemical Stoichiometry. Stoichiometry. Molecular Weight Molecular Weight = Σ Atomic Weights. Stoichiometry. The Mole Concept 1 mole = mass of substance containing 1 Avogadro’s Number of particles. 1 mole = 1 formula weight of substance = 6.02 x 10 23 particles. Stoichiometry.

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Chemical Stoichiometry

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  1. Chemical Stoichiometry

  2. Stoichiometry • Molecular Weight Molecular Weight = Σ Atomic Weights

  3. Stoichiometry • The Mole Concept 1 mole = mass of substance containing 1 Avogadro’s Number of particles. 1 mole = 1 formula weight of substance = 6.02 x 1023 particles

  4. Stoichiometry 1 mole element = atomic weight 1 mole compound = molecular weight 1 mole aluminum = 26.98 amu = 26.98 grams 1 mole NaCl = 58.5 amu = 58.5 grams

  5. Stoichiometry • Conversions Mass to moles => Divide by formula weight Moles to mass => Multiply by formula weight

  6. Stoichiometry • Given 25 grams KBr, convert to moles. (f.wt. KBr = 119 g/mole) Moles KBr = (25 gms / 119 gms/mole ) = 0.2100 mole

  7. Stoichiometry • % Elemental Composition %Element = (TTL Mass Element/Molecular Weight)100% H2SO4 ( f.wt. = 98 amu) %H = [2H/ H2SO4]x100% = [2(1)/98]x100% = 2.04% %S = [1S/ H2SO4]x100% = [1(32)/98]x100% = 32.65% %O = [4O/ H2SO4]x100% = [4(16)/98]x100% = 65.31% Σ % = 2.04% + 32.65% + 65.31% = 100%

  8. Stoichiometry • Empirical and Molecular Formula • Empirical Formula ≡ Smallest whole number ratio of elements in chemical formula • Molecular Formula ≡ Actual whole number ratio of elements in chemical formula Molecular Weight = N • Empirical Formula Wt

  9. Stoichiometry • Determination of Empirical Formula • % => grams => moles => ratio => reduce ratio => Empirical Ratio => Empirical Formula

  10. Stoichiometry • Quantitative Relations in Chemical Equations N2(g) + 3H2(g) => 2NH3(g)

  11. Stoichiometry 1. Given the following equation: 2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O, show what the following molar ratios should be. a. C4H10 / O2 b. O2 / CO2 c. O2 / H2O d. C4H10 / CO2 e. C4H10 / H2O 2. Given the following equation: 2 KClO3 ---> 2 KCl + 3 O2 How many moles of O2 can be produced by letting 12.00 moles of KClO3 react? 3. Given the following equation: 2 K + Cl2 ---> 2 KCl How many grams of KCl is produced from 2.50 g of K and excess Cl2. From 1.00 g of Cl2 and excess K? 4. Given the following equation: Na2O + H2O ---> 2 NaOH How many grams of NaOH is produced from 1.20 x 102 grams of Na2O? How many grams of Na2O are required to produce 1.60 x 102 grams of NaOH?

  12. Stoichiometry 5. Given the following equation: 8 Fe + S8 ---> 8 FeS What mass of iron is needed to react with 16.0 grams of sulfur? How many grams of FeS are produced? 6. Given the following equation: 2 NaClO3 ---> 2 NaCl + 3 O2 12.00 moles of NaClO3 will produce how many grams of O2? How many grams of NaCl are produced when 80.0 grams of O2 are produced? 7. Given the following equation: Cu + 2 AgNO3 ---> Cu(NO3)2 + 2 Ag How many moles of Cu are needed to react with 3.50 moles of AgNO3? If 89.5 grams of Ag were produced, how many grams of Cu reacted? 8. Molten iron and carbon monoxide are produced in a blast furnace by the reaction of iron(III) oxide and coke (pure carbon). If 25.0 kilograms of pure Fe2O3 is used, how many kilograms of iron can be produced? The reaction is: Fe2O3 + 3 C ---> 2 Fe + 3 CO 9. The average human requires 120.0 grams of glucose (C6H12O6) per day. How many grams of CO2 (in the photosynthesis reaction) are required for this amount of glucose? The photosynthetic reaction is: 6 CO2 + 6 H2O ---> C6H12O6 + 6 O2 10. Given the reaction: 4 NH3 (g) + 5 O2 (g) ---> 4 NO (g) + 6 H2O (l) When 1.20 mole of ammonia reacts, the total number of moles of products formed is ?

  13. Stoichiometry - Limiting ReactantsTheoritical & % Yield Methane, CH4, burns in oxygen to give carbon dioxide and water according to the following equation: CH4  + 2 O2 ------>  CO2 + 2 H2O In one experiment, a mixture of 0.250 mol of methane was burned in 1.25 mol of oxygen in a sealed steel vessel.  Find the limiting reactant, if any, and calculate the theoretical yield, (in moles) of water. Chloroform, CHCl3, reacts with chlorine, Cl2, to formcarbon tetrachloride, CCl4, and hydrogen chloride, HCl.  In an experiment 25 grams of chloroform and 25 grams of chlorine were mixed.  Which is the limiting reactant?  What is the maximum yield of CCl4 in moles and in grams?

  14. StoichiometrySolution Phase Reactions • Solution Concentration – Molarity • Molarity(M) = ( Moles Solute / Liter Solution) • Moles Solute = Molarity x Volume(L) • Solution Data ≡ Concentration (Molarity) together with Volume Convert ‘Solution Data’ to moles, solve equation by proportionality.

  15. StoichiometrySolution Phase Reactions • What volume of 0.250M HNO3 (nitric acid) reacts with 44.8-ml of 0.150M Na2CO3(sodium carbonate) in the following reaction? 2HNO3(aq) + Na2CO3(aq) -> 2NaNO3(aq) + H2O(l) + CO2(g) • A solution of 35-ml of 0.25M HCl reacts with 40-ml of 0.30M NaOH according to the following reaction: HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l) Which of the reacting reagents is the limiting reactant and which determines the acidity or alkalinity of the final solution mix?

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