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Baking Soda and Vinegar Limiting Reactant Lab

Baking Soda and Vinegar Limiting Reactant Lab. CH 3 COOH + NaHCO 3  NaCH 3 COO + CO 2 + H 2 O. PPT Init: 1/24/2011 by Daniel R. Barnes.

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Baking Soda and Vinegar Limiting Reactant Lab

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  1. Baking Soda and Vinegar Limiting Reactant Lab CH3COOH + NaHCO3 NaCH3COO + CO2 + H2O PPT Init: 1/24/2011 by Daniel R. Barnes WARNING: As with all my power points, this one contains graphical images and/or other content taken from the world wide web without the permission of the owners of that intellectual property. Please do not copy or distribute this presentation. Its very existence may be illegal.

  2. SWBAT . . . . . . predict how many grams of carbon dioxide should be produced from a given number of grams of baking soda. . . . use the idea of “limiting reactant” to explain the amounts of carbon dioxide produced by various amounts of baking soda reacting with a fixed amount of vinegar.

  3. THE REACTION CH3COOH + NaHCO3 NaCH3COO + CO2 + H2O carbon dioxide acetic acid sodium hydrogen carbonate water sodium acetate

  4. THE REACTION CH3COOH + NaHCO3 NaCH3COO + CO2 + H2O H H Na H C O C H H H H O O H H C C O C C O O O O O O H Na

  5. THE REACTION CH3COOH + NaHCO3 NaCH3COO + CO2 + H2O H Na H C H H O O C C O O O H Let’s animate this cartoon!

  6. MASS MEASUREMENTS VOCABULARY FRONT-LOAD “tare mass” = mass of empty container “net mass” = mass of what’s in the container “gross mass” = mass of container + contents gross = net + tare gross = contents + container net = gross - tare

  7. Stabby Flakes Stabby Flakes NET WT 3 LB NET WT 3 LB Empty box = 0.25 pounds Cereal = 3 pounds Box full of cereal = 3.25 pounds

  8. Stabby Flakes Stabby Flakes + TARE ? = NET GROSS NET WT 3 LB NET WT 3 LB Empty box = 0.25 pounds Cereal = 3 pounds Box full of cereal = 3.25 pounds

  9. MASS MEASUREMENTS NOTE: The scale pictured here is a digital scale, but, as of this writing, the lab worksheet says you’re supposed to use a triple beam balance. Sorry. TBB’s are too hard to draw. Bear with me. When you have the 600 mL beaker and the 50 mL beaker on the scale and they’re both empty, you’ve got a certain amount of glass on the scale. Mass of emtpy container(s) = “tare” mass 251.4 g

  10. MASS MEASUREMENTS “net” mass If you pour exactly 100 g of vinegar in the 600 mL beaker, If you pour exactly 100 g of vinegar in the 600 mL beaker, your new total mass should be exactly 100 g higher than the “tare” mass. 251.4 351.4 g

  11. MASS MEASUREMENTS “net” mass If you were to put exactly five grams of baking soda in the 50 mL beaker, that should bring the new gross mass up to . . . . . . 356.4 g 351.4 356.4 g

  12. MASS MEASUREMENTS If you dump the baking soda into the vinegar . . . 356.4 g

  13. MASS MEASUREMENTS If you dump the baking soda into the vinegar . . . The chemicals react to form bubbles of carbon dioxide 356.4 g

  14. MASS MEASUREMENTS When the bubbles rose and popped, carbon dioxide escaped from the 600 mL beaker and into the air. Therefore, the gross mass went . . . CO2 CO2 CO2 . . . down, in this case by 2.6 grams. 356.4 353.8 g

  15. MASS MEASUREMENTS 356.4 g = gross mass before the reaction - 353.8 g = gross mass after the reaction 2.6 g = amount of CO2 produced We assume that however much the mass went down . . . . . . is the amount of CO2 produced. 353.8 g

  16. MASS MEASUREMENTS 356.4 g = gross mass before the reaction - 353.8 g = gross mass after the reaction 2.6 g = amount of CO2 produced The law of conservation of matter says that . . . . . . matter can’t be destroyed during a chemical reaction. The drop in mass doesn’t represent matter that was destroyed. It represents matter that went into the air. 353.8 g

  17. Stoichiometry “that portion of chemistry dealing with numerical relationships in chemical reactions; the calculation of quantities of substances invovled in chemical reactions.” ~Prentice Hall Chemistry, glossary, pg R117 That’s what your textbook says, but in this class, “stoichiometry” pretty much means . . . “the process of calculating how many grams of one chemical are consumed/produced during a reaction when the number of grams of any other reactant/product is given.” That’s what you’re going to have to be able to do on the CST, so that’s what you’re going to have to be able to do on my tests, too. Anything else is extra credit.

  18. Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below: g 8.9 17g CH3COOH + NaHCO3 NaCH3COO + CO2 + H2O 17g NaHCO3 1 mol NaHCO3 1 mol CO2 44 g CO2 x x x 1 g NaHCO3 1 mol NaHCO3 1 mol CO2 84 NaHCO3: Na: 1 x 23 = 23 CO2: C: 1 x 12 = 12 O: 2 x 16 = 32 17 x 44 x 1 = 1 1 H: 8.9 ) = 12 x 12 1 84 748 C: 748 x 16 = 48 3 O: 44 g/mol 84 g/mol

  19. Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below: g 8.9 17g CH3COOH + NaHCO3 NaCH3COO + CO2 + H2O 17g NaHCO3 1 mol NaHCO3 1 mol CO2 44 g CO2 x x x 1 g NaHCO3 1 mol NaHCO3 1 mol CO2 84 The mass of the known substance goes on the top of the first fraction. NaHCO3: Na: 1 x 23 = 23 CO2: C: 1 x 12 = 12 O: 2 x 16 = 32 17 x 44 x 1 = 1 1 H: 8.9 ) = 12 x 12 1 84 748 C: 748 x 16 = 48 3 O: 44 g/mol 84 g/mol

  20. Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below: g 8.9 17g CH3COOH + NaHCO3 NaCH3COO + CO2 + H2O 17g NaHCO3 1 mol NaHCO3 1 mol CO2 44 g CO2 x x x 1 g NaHCO3 1 mol NaHCO3 1 mol CO2 84 The molar mass of the known substance goes on the bottom of the second fraction. NaHCO3: Na: 1 x 23 = 23 CO2: C: 1 x 12 = 12 O: 2 x 16 = 32 17 x 44 x 1 = 1 1 H: 8.9 ) = 12 x 12 1 84 748 C: 748 x 16 = 48 3 O: 44 g/mol 84 g/mol

  21. Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below: g 8.9 17g CH3COOH + NaHCO3 NaCH3COO + CO2 + H2O 17g NaHCO3 1 mol NaHCO3 1 mol CO2 44 g CO2 x x x 1 g NaHCO3 1 mol NaHCO3 1 mol CO2 84 The coefficients go in the third fraction, known as the “mole ratio” fraction. NaHCO3: Na: 1 x 23 = 23 CO2: C: 1 x 12 = 12 O: 2 x 16 = 32 17 x 44 x 1 = 1 1 H: 8.9 ) = 12 x 12 1 84 748 C: 748 x 16 = 48 3 O: 44 g/mol 84 g/mol

  22. Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below: g 8.9 17g CH3COOH + NaHCO3 NaCH3COO + CO2 + H2O 17g NaHCO3 1 mol NaHCO3 1 mol CO2 44 g CO2 x x x 1 g NaHCO3 1 mol NaHCO3 1 mol CO2 84 Yeah. I know. There are no coeffcients in this equation. That’s why I put 1’s in the fraction. NaHCO3: Na: 1 x 23 = 23 CO2: C: 1 x 12 = 12 O: 2 x 16 = 32 17 x 44 x 1 = 1 1 H: 8.9 ) = 12 x 12 1 84 748 C: 748 x 16 = 48 3 O: 44 g/mol 84 g/mol

  23. Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below: g 8.9 17g CH3COOH + NaHCO3 NaCH3COO + CO2 + H2O 17g NaHCO3 1 mol NaHCO3 1 mol CO2 44 g CO2 x x x 1 g NaHCO3 1 mol NaHCO3 1 mol CO2 84 The molar mass of the unknown substance goes on the top of the last fraction. NaHCO3: Na: 1 x 23 = 23 CO2: C: 1 x 12 = 12 O: 2 x 16 = 32 17 x 44 x 1 = 1 1 H: 8.9 ) = 12 x 12 1 84 748 C: 748 x 16 = 48 3 O: 44 g/mol 84 g/mol

  24. Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below: g 17g CH3COOH + NaHCO3 NaCH3COO + CO2 + H2O 17g NaHCO3 1 mol NaHCO3 1 mol CO2 44 g CO2 x x x 1 g NaHCO3 1 mol NaHCO3 1 mol CO2 84 Okay. Now you do the same calculation, but for the 2 masses of baking soda your group used. NaHCO3: Na: 1 x 23 = 23 CO2: C: 1 x 12 = 12 O: 2 x 16 = 32 17 x 44 x 1 = 1 1 H: 8.9 ) = 12 x 12 1 84 748 C: 748 x 16 = 48 3 O: 44 g/mol 84 g/mol

  25. Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below: g 17g CH3COOH + NaHCO3 NaCH3COO + CO2 + H2O 17g NaHCO3 1 mol NaHCO3 1 mol CO2 44 g CO2 x x x 1 g NaHCO3 1 mol NaHCO3 1 mol CO2 84 I need to see the four-fraction chain for each calculation . . . NaHCO3: Na: 1 x 23 = 23 CO2: C: 1 x 12 = 12 O: 2 x 16 = 32 17 x 44 x 1 = 1 1 H: 8.9 ) = 12 x 12 1 84 748 C: 748 x 16 = 48 3 O: 44 g/mol 84 g/mol

  26. Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below: g 17g CH3COOH + NaHCO3 NaCH3COO + CO2 + H2O 17g NaHCO3 1 mol NaHCO3 1 mol CO2 44 g CO2 x x x 1 g NaHCO3 1 mol NaHCO3 1 mol CO2 84 . . . and, somewhere on your paper, I need to see the molar mass calculations for NaHCO3 & CO2. NaHCO3: Na: 1 x 23 = 23 CO2: C: 1 x 12 = 12 O: 2 x 16 = 32 17 x 44 x 1 = 1 1 H: 8.9 ) = 12 x 12 1 84 748 C: 748 x 16 = 48 3 O: 44 g/mol 84 g/mol

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