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Stoichiometry

Stoichiometry. Stoichiometry is the part of chemistry that studies amounts of substances that are involved in reactions . It could be amounts of substances before the reaction or amount of material that is produced by the reaction. Stoichiometry is all about amounts.

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Stoichiometry

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  1. Stoichiometry

  2. Stoichiometry is the part of chemistry that studies amounts of substances that are involved in reactions. • It could be amounts of substances before the reaction or amount of material that is produced by the reaction. • Stoichiometry is all about amounts.

  3. What a balanced chemical equation tells us • A chemical equation represents the mole ratio of reactants and products • CuCl2 + 2AgNO3 2AgCl + Cu(NO3)2 1mol 2mol 2mol 1mol • NaOH + HCl  NaCl + H2O 1mol 1mol 1mol 1mol • 2CO + O2 -> 2CO2 2mol 1mol 2mol • Mg + 2HCl -> MgCl2 + H2 1 mol 2 mol 1 mol 1 mol

  4. Stoichiometry will tell you that if you have ten million atoms of sodium (Na) and only one atom of chlorine (Cl) you can only make one molecule of sodium chloride (NaCl). Nothing you can do will change that. Like this: • 10,000,000 Na + 1 Cl --> NaCl + 9,999,999 Na

  5. Limiting reagent • A reagent is another word for reactant • A reaction will stop when one reactant is used up before the other • This is called a limiting reagent • The other reactant is the excess reagent • Blue is the limiting reagent • Red is the excess reagent

  6. Limiting reagent 2H2 + O2 2H2O Initial mol 8 6 0 Reaction mol 8 4 8 Final mol 0 2 8

  7. Mass-Mass stoichemtry • Involves solving a problem in which the mass of a reactant or a product is given

  8. Calculating the mass of a substance given the mass of another reactant or product • Write a balanced chemical equation • Identify known and unknown quantities of substances • Calculate the number of moles of known quantities • Find the molar ratio and use this to calculate the number of moles of a required substance Mol of required substance = Molar ratio Mol of given substance 5. Calculate the quantity o the required, unknown, substance

  9. Calculations • Calculate the number of moles of CO2 formed in the combustion of ethane C2H6 in a process when 35.0 mol of O2 is consumed.The reaction is: • 2 C2H6 + 7 O2 4 CO2 + 6 H2O • 4 CO2 : 7 O2 • 35.0 mol O2 x 4 CO2 mol = 20.0 mol CO2 • 7 mol O2 mol

  10. Two moles of Mg and five moles of O2 are placed in a reaction vessel, and then the Mg is ignited according to the reaction Mg + O2 MgO. Identify the limiting reagent in this experiment. • 2 Mg + O2 2 MgO 2 mol 1mol (I have 5, 5 -1 = 4 mol unused) • Four moles of oxygen will remain unreacted. • Oxygen is the excess reagent, and Mg is the limiting reagent.

  11. What mass of iron (III) oxide is formed from the complete combustion of 183.5g of pyites (FeS2) • FeS2 + O2 Fe2O3 + SO2 • 4FeS2 + 11O2 2Fe2O3 + 8SO2 • molar mass of FeS2 = 55.85 + (2 x 32.1) = 120.05g/mol = 183.5g x 1 mol 120.05g = 1.529 mol

  12. 3. Find the molar ratio of 4FeS2 to 2Fe2O3 4:2 2:1 Mol of Fe2O3 = 2 FeS2 = 1 4 2 = 1.529 mol 2 Mol of Fe2O3 = 0.7643 mol 4. Molar mass of Fe2O3 = 2 x 55.85 + 3 x 16 = 159.7g/mol 0.7643 mol x 159.7g 1mol = 122.1g of Fe2O3

  13. Calculate the mass of water produced when 2.8g of CH4 is burnt in the air • CH4 + 2O2 2H2O + CO2 • Molar mass = 12 + 4 = 16g/mol Mol of CH4 = 2.8g x 1 mol 16g Mol of CH4 = 0.175 mol 3. Molar ratio CH4 : H2O 1:2 CH4 = 1 = 0.175 mol = 0.0875 mol of H2O H2O 2 2 4. Molar mass of H20 = 2+16 = 18g/mol Mass of H20 = 0.0875 mol x 18g = 1.575g 1 mol

  14. Limiting reagent • How much precipitate can you make with only 2.6mol of KCl? • AgNO3 + KCl  • AgNO3(aq) + KCl(aq) AgCl (s) + KNO3(aq) 2.6mol • KCL is the limiting reagent we only have a certain amount of it

  15. Balance equation AgNO3(aq) + KCl(aq) AgCl (s) + KNO3(aq) • Determine molar ratio • KCl to AgCl is 1:1 I have 2.6mol of KCl I have 2.6 mol of AgCl Molar mass of AgCl = 143.32g/mol 2.6mol KCl x 1mol of AgCl x 143.32g of AgCl 1 mol of KCL 1mol of AgCl = 372.63g of AgCl

  16. If 6.3g of S8 react with Pb, how many grams of PbS are formed? • Pb + S8 PbS • 8Pb + S8 8PbS 6.3g ? Molar mass of S8 = 32.06 x 8 = 256.48g/mol Mol of S8 = 6.6g x 1mol 256.48g = 0.025 mol Molar ratio = PbS 8 S8 1 Mol of PbS = 8 x 0.025 = 1.647 mol Molar mass of PbS = 239.27g/mol Mass of PbS = 1.647mol x 239.27g 1 mol = 394.06 g

  17. Determine the limiting reagent • What mass of liquid water is formed when 2.3g of H2 gas and 4.55g of O2 gas react together • 2H2 + O2 2H2O • 2.3g 4.55g Molar mass of H2 = 2g/mol Mol of H2 = 2.3g x 1mol = 1.15mol 2g Molar mass of 02 = 32g/mol Mol of O2 = 4.55 x 1 mol = 0.142mol 32g

  18. 2H2 + O2 2H2O 1.15 0.142 0.142 x 2 = 0.284mol of H2 Which is the limiting reagent? 02 2H2 + O2 2H2O 0.284 0.142 0.284 Molar mass of H2O = 18g/mol Mass of H2O = 0.282mol x 18g 1 mol = 5.076g

  19. What volume of 0.100mol/L H2SO4 acid reacts completely with 17.8mL of 0.15 of potassium hydroxide? Acid + base  salt + water KOH + H2SO4 K2SO4 + H20 2KOH + H2SO4 K2SO4 + 2H20 0.15M 0.1M 17.8mL ? (0.0178L) Mol of KOH = 0.0178L x 0.15mol 1L = 0.00267mol Molar ratio = H2SO4 1 KOH 2 Mol of H2SO4 = ½ x 0.00267mol L of H2SO4 = 0.00136mol x 1L 0.1mol L of H2SO4 = 0.01335L

  20. Zinc metal is reacted with 400mL of a 0.25mol/L solution of sulfuric acid. Calculate the mass of zinc sulfate formed

  21. Questions • Pg 263 Q 1, 2, 4, 6,7,8 • Pg 267 Q 9

  22. Titration • A titration is a method of analysis that will allow you to determine the precise endpoint of a reaction and therefore the precise quantity of reactant in the titration flask. A buret is used to deliver the second reactant to the flask and an indicator or pH Meter is used to detect the endpoint of the reaction.

  23. http://www.wesleylearning.ie/resources/science/chemistry/experiments/ethanoic_acid_vinegar/index.htmhttp://www.wesleylearning.ie/resources/science/chemistry/experiments/ethanoic_acid_vinegar/index.htm

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