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Balancing and stoichiometry Stoichiometry uses balanced equations to make predictions of products or reactants. It fol

Balancing and stoichiometry Stoichiometry uses balanced equations to make predictions of products or reactants. It follows the laws of mass conservation . For example: Let’s look at substance A and B that react to yield substance C and D. A + B → C + D.

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Balancing and stoichiometry Stoichiometry uses balanced equations to make predictions of products or reactants. It fol

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  1. Balancing and stoichiometry Stoichiometry uses balanced equations to make predictions of products or reactants. It follows the laws of mass conservation. For example: Let’s look at substance A and B that react to yield substance C and D. A + B → C + D If this equation respects the law of mass conservation, then, the total mass of A and B should equal to the total mass of C and D. A + B →C + D 30g 20g 27g 23g 50 g 50 g Thus, a conservation of matter where the total masses are equal.
  2. Now, in the previous example, it was all a 1:1 ratio, i.e. 1A, 1 B, 1 C and 1D. But in reality you don’t usually have this scenario. This is where stoichiometry comes in handy at predicting masses or numbers of moles. Let’s look at the following example: Ca(OH)2(aq)+ H3P(aq)→ H2O(l) + Ca3P2 (aq) Initially, lets’ balance this equation. Hydrogen and oxygen are usually balanced at the end, you would start with the heavier atoms. Also, notice the subscripts: (aq): aqueous, dissolved in water (l): liquid You can also have (g) which is gas and (s) which is solid.
  3. Ca(OH)2(aq)+ H3P(aq)→ H2O(l) + Ca3P2 (aq)
  4. 3 Ca(OH)2(aq)+ H3P(aq)→ H2O(l) + Ca3P2 (aq)
  5. 3 Ca(OH)2(aq)+ 2H3P(aq)→ H2O(l) + Ca3P2 (aq)
  6. 3 Ca(OH)2(aq) + 2 H3P(aq) → 6H2O(l) + Ca3P2(aq) Now, all is balanced. Let’s do some stoichiometry.
  7. What number of moles of Ca(OH)2 do you need to react with 5 moles of H3P? . 3 Ca(OH)2(aq) + 2 H3P(aq)→ 6H2O(l) + Ca3P2(aq)
  8. In this step, you will cross multiply.. 3 Ca(OH)2(aq) + 2 H3P(aq)→ 6H2O(l) + Ca3P2(aq) So, it takes 7.5 moles of Ca(OH)2 to react with 5 moles of H3P.
  9. Let’s look at another example... CH4(g) + O2(g) → CO2(g) + H2O(l) First, balance. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
  10. How many moles of oxygen (O2) are needed to produce 40 grams of CO2? CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
  11. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) In this case, we have grams. You can not cross-multiply grams with moles. So the grams must be converted to moles. We use proportions or use n = m/mMm. 1 mol of CO2 → 44 g x → 40 g
  12. 1 mol of CO2 → 44 g x → 40 g x = 40 / 44 = 0.91 mol
  13. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) We can now cross-multiply..
  14. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) So, it takes 1.82 moles of O2 to react with 0.91 moles of CO2 or 40 g of CO2..
  15. Now, let’s look at an example that uses volume and concentration. For sake of simplicity, we’ll use the former example of Ca(OH)2.
  16. Calculate the number of moles of H3P needed to react with Ca(OH)2 if you are given 200 mL (0,2 L)of a Ca(OH)2 solution concentrated at 2.25 mol/L or 2.25 M. 3 Ca(OH)2(aq) + 2 H3P(aq)→ 6H2O(l) + Ca3P2(aq) So, we need to calculate the number of moles of Ca(OH)2 used. To do so, we’ll use C = n/V formulae to calculate the number of moles..
  17. n C x V According to the triangle, n = C x V.. n = C x V = 2.25 mol x 0.2 L L = 0.45 mol
  18. 3 Ca(OH)2(aq) + 2 H3P(aq)→ 6H2O(l) + Ca3P2(aq) Now, cross multiply... So, it takes 0.3 moles of H3P to react with 0.45 moles of Ca(OH)2.
  19. Precipitate and stoichiometry A precipitate is a solid water-insoluble substance that results from the reaction of two aqueous ions. The following table shows some examples of combinations of ions that result in precipitates.
  20. So, as you can see in the diagram below, you add two reactants together to form a precipitate and a another soluble substance. → + Reactant A dissolved in water Reactant B dissolved in water In water
  21. Let’s look at an example. When reacting sodium iodide (NaI) with mercury(II) chloride (HgCl2), a precipitate forms. To find out which compound will become the precipitate, you must first learn how to create product formulas, let’s look at a simple example. AB+ CD→AD +CB Notice in this example how the atoms switched places.. AB+ CD→AD +CB Let’s go back to sodium iodide (NaI) and mercury(II) chloride (HgCl2).
  22. NaI(aq) + HgCl2(aq)→ ?+ ? Sodium iodide.. Mercruy (II) chloride.. Both substances are aqueous. There will be a solid (s) precipitate and the other compound will be aqueous. So, now let’s switch the atoms as we did before. AB + CD→ AD + CB NaI(aq) + HgCl2(aq)→ HgI2 + NaCl Notice that the metals (Na and Hg) always go first..But now, which product will be the precipitate? To find the answer to that question, consult your table of precipitates in the next slide.
  23. No precipitate forms Precipitate forms As you can see mercury with iodine forms a precipitate and sodium with chlorine does not.
  24. So, the equation is the following: NaI(aq) + HgCl2(aq)→ HgI2(s) +NaCl(aq) Animated gif of the reaction foNaIwith HgCl2. Notice the reddish precipitate that forms in this reaction. With time, that reddish powder will settle at the bottom of the beaker.
  25. Now, let’s do an exercise with the previous equation. But first let’s balance.: 2NaI(aq) + HgCl2(aq)→ HgI2(s) +2NaCl(aq)
  26. What mass of salt precipitate can you form from the chemical reaction between 30 mL of 1.0M NaIand 15 mL of 1.0M HgCl2 salt solutions? 2) Is there conservation of matter in this reaction? Now, to solve number 1, we need to do a bit of stoichiometry. 2NaI(aq) + HgCl2(aq) → HgI2(s) +2NaCl(aq) So we need to find the number of moles for both reactants. We use C= n/V.
  27. n C x V According to the triangle, n = C x V.. For NaI: n = C x V = 1 mol x 0.03 L L = 0.03 mol
  28. n C x V According to the triangle, n = C x V.. For HgCl2: n = C x V = 1 mol x 0.015 L L = 0.015 mol
  29. 2NaI(aq) + HgCl2(aq) → HgI2(s) +2NaCl(aq) Now, to figure out the precipitate, you do stoichiometry with cross-multiplication. You can also figure out the amount of NaCl produced too. We’ll use the 0.015 of HgCl2 to solve.
  30. 2NaI(aq) + HgCl2(aq) → HgI2(s) +2NaCl(aq) 2NaI(aq) + HgCl2(aq) → HgI2(s) +2NaCl(aq)
  31. So now, this is what we have: 2NaI(aq) + HgCl2(aq) → HgI2(s) +2NaCl(aq) To find the mass of the precipitate, you can use n = m/mm or proportions.
  32. 1 mol of HgI2 → 455 g 0.015 mole → x x = 0.015 * 455 = 6.83 g So, the answer to the first question is 6.83 g of precipitate (HgI2) are produced form the reaction of NaI with HgCl2
  33. To answer the second question, just convert all reactants and product into grams to see if they follow the law of mass conservation. 2NaI(aq) + HgCl2(aq) → HgI2(s) +2 NaCl(aq) Yes, there is mass conservation.
  34. So, this is what the reaction appears to be: → + HgI2(s) HgCl2(aq) NaI(aq) NaCl(aq) 15 mL 30 mL
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