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STOICHIOMETRY

STOICHIOMETRY. What is stoichiometry?. Stoichiometry is the quantitative study of reactants and products in a chemical reaction. What do you need?. You will need to use Balanced chemical equation Ratios from the coefficients molar masses

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STOICHIOMETRY

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  1. STOICHIOMETRY

  2. What is stoichiometry? • Stoichiometry is the quantitative study of reactants and products in a chemical reaction.

  3. What do you need? You will need to use • Balanced chemical equation • Ratios from the coefficients • molar masses Note: This type of problem is often called "mass-mass."

  4. Mass-MassProblems There are four steps involved in solving these problems: • Make sure you are working with a properly balanced equation. • Convert grams of the substance given in the problem to moles. • Use the mole ratio • Convert moles of the substance just solved for into grams.. Or whatever the problem asks for.

  5. Mole Ratios A mole ratio converts moles of one compound in a balanced chemical equation into moles of another compound.

  6. Example Reaction between magnesium and oxygen to form magnesium oxide. 2 Mg(s) + O2(g) 2 MgO(s) Mole Ratios: 2 : 1 : 2

  7. If 4.0 moles of oxygen react, find the moles of MgO produced. Reaction between magnesium and oxygen to form magnesium oxide. 2 Mg(s) + O2(g) 2 MgO(s) Mole Ratios: 2 : 1 : 2 4.0X Answer: 8.0 moles of MgO

  8. More Practice A can of butane lighter fluid contains 1.20 moles of butane (C4H10). Calculate the number of moles of carbon dioxide given off when this butane is burned. Balance the equation: C4H10 + O2 CO2 + H2O

  9. Mole-Mole Problems The balanced chemical equation for the reaction 2 C4H10 + 13 O2 8 CO2 + 10 H2O Mole ratios: C4H10 CO2 2 : 8 [ coefficients] 1.2 : X [ problem] By cross-multiplication, X = 4.8 moles of CO2 given off

  10. Practice Problems 1) N2 + 3 H2 ---> 2 NH3 Write the mole ratios for N2 to H2 and NH3 to H2. N2 + 3 H2 ---> 2 NH3 1 to 3 to 2

  11. N2 (g) + 3 H2 (g) ---> 2 NH3 (g) • If 0.2 moles of nitrogen react, find the moles of ammonia produced • If 0.5 moles of hydrogen react, find the moles of ammonia produced. Answers: 0.4 moles of ammonia 0.33 moles of ammonia

  12. Mole-Mass Problems • 1.50 mol of KClO3 decomposes. How many grams of O2 will be produced? [K = 39.0, Cl = 35.5, O = 16.0] 2 KClO3 2 KCl + 3 O2

  13. Example 2 KClO3 2 KCl + 3 O2 2 : 3 1.50 : X Solve for X = 2.25 mol Convert to mass 2.25 mol x 32.0 g/mol = 72.0 grams

  14. 2 KClO3 2 KCl + 3 O2 • In order to produce 2.75 mol of KCl. How many grams of KClO3 would be required? Solution: KClO3 : KCl 2 : 2 (coefficients) X : 2.75 (from the problem) Solve for X = 2.75 mol Find the mass: 2.75 mol X 122.55 g/mol = 337 grams

  15. Molarvolume The molar volume is the volume occupied by one mole of ideal gas at STP. Its value is: 22.4 dm3

  16. Conversion of mole to volume at STP conditions: moles of gas = Volume Molar volume Note: STP conditions are standard temperature and pressure. 0 oC and 1 atmosphere

  17. CaCO3(s) CaO(s) + CO2(g) Calculate the volume of carbon dioxide formed at STP in ‘dm3' by the complete thermal decomposition of 3.125 g of pure calcium carbonate (Relative atomic mass of Ca=40.0, C=12.0, O=16.0) Solution: Convert the mass to moles: The molar mass of CaCO3 = 40.0 + 12.0 + (16.0 x 3) = 100.0 g mol-1 Mole = mass/molar mass 3.125/100.0 = 0.03125mol

  18. Practice Problems Calcium carbonate decomposes: Mole ratio 1 : 1 problem 0.03125 : X X = 0.03125 mole of CO2 Volume = 0.03125 mole x 22.4dm3 =0.70 dm3

  19. Theoretical Yield • These calculations are all theoretical. • If the reaction worked perfectly, this is how much substance would be produced. • In the laboratory, the experimental yield is nearly always less than the theoretical yield.

  20. Percent yield The percent of the compound produced in the experiment. Experimental yield x 100 = percent yield Theoretical yield

  21. If the percent yield is… • LESS THAN than 100% The reaction did not go to completion Some of the product was lost during purification steps. • More than 100% The product is not pure. There are other compounds contaminating the sample.

  22. Calculate the percent yield 2.43 grams of magnesium reacts with excess oxygen to form magnesium oxide. In the lab, 3.00 grams of MgO was isolated. Calculate the percent yield.

  23. 2 Mg(s) + O2(g)  2 MgO (s) • Find moles of Mg: 2.43 grams/24.3 = 0.10 moles Mg Find moles of MgO: The ratio is 2 to 2, so 0.10 moles MgO form Find theoretical mass of MgO: 0.10 x 40.3 = 4.03 grams

  24. 2 Mg(s) + O2(g)  2 MgO (s) • Percent yield = Experimental x 100 = Theoretical 3.00 grams x 100 = 75% 4.03 grams

  25. This powerpoint was kindly donated to www.worldofteaching.com http://www.worldofteaching.com is home to over a thousand powerpoints submitted by teachers. This is a completely free site and requires no registration. Please visit and I hope it will help in your teaching.

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