Stoichiometry ok1294993172

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Stoichiometry ok1294993172

  1. 1. STOICHIOMETRY
  2. 2. What is stoichiometry? <ul><li>Stoichiometry is the quantitative study of reactants and products in a chemical reaction. </li></ul>
  3. 3. What You Should Expect <ul><li>Given : Amount of reactants </li></ul><ul><li>Question: how much of products can be formed. </li></ul><ul><li>Example </li></ul><ul><li>2 A + 2B 3C </li></ul><ul><li>Given 20.0 grams of A and sufficient B, how many grams of C can be produced? </li></ul>
  4. 4. What do you need? <ul><li>You will need to use </li></ul><ul><li>molar ratios, </li></ul><ul><li>molar masses, </li></ul><ul><li>balancing and interpreting equations, and </li></ul><ul><li>conversions between grams and moles. </li></ul><ul><li>Note: This type of problem is often called &quot;mass-mass.&quot; </li></ul>
  5. 5. Steps Involved in Solving Mass-Mass Stoichiometry Problems <ul><li>Balance the chemical equation correctly </li></ul><ul><li>Using the molar mass of the given substance, convert the mass given to moles. </li></ul><ul><li>Construct a molar proportion (two molar ratios set equal to each other) </li></ul><ul><li>Using the molar mass of the unknown substance, convert the moles just calculated to mass. </li></ul>
  6. 6. Mole Ratios <ul><li> A mole ratio converts moles of one compound in a balanced chemical equation into moles of another compound. </li></ul>
  7. 7. Example <ul><li>Reaction between magnesium and oxygen to form magnesium oxide. ( fireworks) </li></ul><ul><li>2 Mg(s) + O2(g) 2 MgO(s) </li></ul><ul><li>Mole Ratios: </li></ul><ul><li>2 : 1 : 2 </li></ul>
  8. 8. Practice Problems <ul><li>1) N 2 + 3 H 2 ---> 2 NH 3 </li></ul><ul><li>Write the mole ratios for N 2 to H 2 and NH 3 to H 2 . </li></ul><ul><li>2) A can of butane lighter fluid contains 1.20 moles of butane (C 4 H 10 ). Calculate the number of moles of carbon dioxide given off when this butane is burned. </li></ul>
  9. 9. Mole-Mole Problems <ul><li>Using the practice question 2) above: </li></ul><ul><li>Equation of reaction </li></ul><ul><li>2C 4 H 10 + 13O 2 8CO 2 + 10H 2 O </li></ul><ul><li>Mole ratio </li></ul><ul><li>C 4 H 10 CO 2 </li></ul><ul><li> 1 : 4 [ bases] </li></ul><ul><li> 1.2 : X [ problem] </li></ul><ul><li>By cross-multiplication, X = 4.8 mols of CO 2 given off </li></ul>
  10. 10. Mole-Mass Problems <ul><li>Problem 1 : 1.50 mol of KClO 3 decomposes. How many grams of O 2 will be produced? [k = 39, Cl = 35.5, O = 16] </li></ul><ul><li>2 KClO 3 2 KCl + 3 O 2 </li></ul>
  11. 11. Three steps…Get Your Correct Answer <ul><li>Use mole ratio </li></ul><ul><li>Get the answer in moles and then </li></ul><ul><li>Convert to Mass. [Simple Arithmetic] </li></ul><ul><li>Hello! </li></ul><ul><li>If you are given a mass in the problem, you will need to convert this to moles first. Ok? </li></ul>
  12. 12. Let’s go! <ul><li>2 KClO 3 2 KCl + 3 O 2 </li></ul><ul><li>2 : 3 </li></ul><ul><li>1.50 : X </li></ul><ul><li>X = 2.25mol </li></ul><ul><li>Convert to mass </li></ul><ul><li>2.25 mol x 32.0 g/mol = 72.0 grams </li></ul><ul><li>Cool! </li></ul>
  13. 13. Try This: <ul><li>We want to produce 2.75 mol of KCl. How many grams of KClO 3 would be required? </li></ul><ul><li>Soln </li></ul><ul><li>KClO 3 : KCl </li></ul><ul><li>2 : 2 </li></ul><ul><li>X : 2.75 </li></ul><ul><li>X = 2.75mol </li></ul><ul><li>In mass: 2.75mol X 122.55 g/mol </li></ul><ul><li>= 337 grams zooo zimple! </li></ul>
  14. 14. Mass-Mass Problems <ul><li>There are four steps involved in solving these problems: </li></ul><ul><li>Make sure you are working with a properly balanced equation. </li></ul><ul><li>Convert grams of the substance given in the problem to moles. </li></ul><ul><li>Construct two ratios - one from the problem and one from the equation and set them equal. Solve for &quot;x,&quot; which is usually found in the ratio from the problem. </li></ul><ul><li>Convert moles of the substance just solved for into grams. </li></ul>
  15. 15. Mass-Volume Problems <ul><li>Just follow mass-mass problem to the penultimate level </li></ul>
  16. 16. Like this: <ul><li>There are four steps involved in solving these problems: </li></ul><ul><li>Make sure you are working with a properly balanced equation. </li></ul><ul><li>Convert grams of the substance given in the problem to moles. </li></ul><ul><li>Construct two ratios - one from the problem and one from the equation and set them equal. Solve for &quot;x,&quot; which is usually found in the ratio from the problem. </li></ul><ul><li>Convert moles of the substance just solved for into Volume. </li></ul>
  17. 17. Conversion of mole to volume <ul><li>No of moles = Volume </li></ul><ul><li>Molar volume </li></ul><ul><li>Can you remember a similar equation? </li></ul>
  18. 18. Molar volume <ul><li>The molar volume is the volume occupied by one mole of ideal gas at STP. Its value is: 22.4dm 3 </li></ul>
  19. 19. Practice Problems <ul><li>Calculate the volume of carbon dioxide formed at STP in ‘dm 3 ' by the complete thermal decomposition of 3.125 g of pure calcium carbonate (Relative atomic mass of Ca=40, C=12, O=16) </li></ul><ul><li>Solution: </li></ul><ul><li>Convert the mass to mole: </li></ul><ul><li>Molar mass of CaCO 3 = 40 + 12 + (16 x 3) = 100gmol -1 </li></ul><ul><li>Mole = mass/molar mass </li></ul><ul><li>3.125/100 = 0.03125mol </li></ul>
  20. 20. Practice Problems <ul><li>As per the equation, </li></ul><ul><li>Mole ratio 1 : 1 </li></ul><ul><li>problem 0.03125mol X </li></ul><ul><ul><ul><ul><li>X = 0.03125mol of CO 2 </li></ul></ul></ul></ul><ul><ul><ul><ul><li>Convert mole to volume [slide 17] </li></ul></ul></ul></ul><ul><ul><ul><ul><li>Volume = ( 0.03125 x 22.4)dm 3 </li></ul></ul></ul></ul><ul><ul><ul><ul><li> = 0.7dm 3 </li></ul></ul></ul></ul>

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