1 / 14

Stoichiometry 3

Stoichiometry 3. Gas and Solution Stoichiometry / Titration. Review. So far we’ve learned that you can convert from moles of one substance to moles of any other substance in one step. Provided you have a balanced chemical equation.

daire
Download Presentation

Stoichiometry 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Stoichiometry 3 Gas and Solution Stoichiometry / Titration

  2. Review • So far we’ve learned that you can convert from moles of one substance to moles of any other substance in one step. • Provided you have a balanced chemical equation. • You can also start stoichiometry problems with grams and particles (atoms, molecules, ions, etc) provided you have the appropriate conversion factors. • 1 mol = 6.022x1023 particles • 1 mol = (molar mass) grams • The molar mass is unique to every substance. • Have to work it out on a case by case basis.

  3. particles A particles A g A g B mol A mol B L A(g) L B(g) L A(aq) L B(aq) Stoichiometry Map 1 mol = 6.022x1023 particles 1 mol = 6.022x1023 particles 1 mol = (molar mass) g 1 mol = (molar mass) g BCE 1 mol = 22.4 L 1 mol = 22.4 L (Molarity) mol = 1 L (Molarity) mol = 1 L

  4. Gas Stoichiometry • How many liters of CO2 (at STP) are produced by the decomposition of 14.0 g CaCO3 in the following reaction? • CaCO3(s)  CaO(s) + CO2(g) • Given: 14.0 g CaCO3 • Want: ??? L CO2 • Conversion factors: • 1 mol CaCO3 = 100.09 g CaCO3 • 1 mol CaCO3 yields 1 mol CO2 • At STP, 1 mol CO2 = 22.4 L CO2 • Plan: 14.0 g CaCO3 mol CaCO3  mol CO2  L CO2 14.0 g CaCO3 = 3.13 L CO2

  5. Gas Stoichiometry • If 25.0 mL of H2 (at STP) are available to react with excess Fe2O3, how many grams of Fe can be produced? • Fe2O3(s) + 3H2(g)  2Fe(s) + 3H2O(l) • Given: 25.0 mL H2 • Want: ??? g Fe • Conversion factors: • 0.001 L = 1 mL • At STP, 1 mol H2 = 22.4 L H2 • 3 mol H2 yield 2 mol Fe • 1 mol Fe = 55.85 g Fe • Plan: 25.0 mL H2 L H2  mol H2  mol Fe  g Fe 25.0 mL H2 = 0.0416 g Fe

  6. Gas Stoichiometry • How to simplify volume-volume problems. • If the given and wanted values are both in liters (or both in milliliters) you can simplify the problem a bit. • Why? • Because you’d have to divide by 22.4 to get from liters to moles, then multiply by 22.4 again to get from moles to liters on the other side. • If you start and end with milliliters, the same thing happens with the mL  L conversion on one side and the L  mL conversion on the other side. • Since those operations cancel each other out, you can bypass them if you’re cautious!

  7. Gas Stoichiometry • What volume of hydrogen gas (at STP) is needed to react with 20.0 L of N2 gas (at STP) in the following reaction? • N2(g) + 3H2(g)  2NH3(g) • Given: 20.0 L N2 • Want: ??? L H2 • Conversion factors: • 1 mol N2 reacts with 3 mol H2 • At STP, 1 mol N2 = 22.4 L N2 • At STP, 1 mol H2 = 22.4 L H2 • Since both moles translate to liters in the same way, we can bypass moles and convert from L N2 L H2 directly. 20.0 L N2 = 60.0 L H2

  8. Solution Stoichiometry • In solution stoichiometry we use the molar concentrations of solutions as conversion factors from liters to moles. • (Molarity) moles = 1 liter • In a 2.50 M HCl solution, there are 2.50 mol HCl dissolved in 1 L of solution.

  9. Solution Stoichiometry • How many grams of aluminum are needed to react with 200. mL of 0.250 M CuCl2? • 2Al(s) + 3CuCl2(aq)  2AlCl3(aq) + 3Cu(s) • Given: 200. mL CuCl2 • Want: ??? g Al • Conversion factors: • 0.001 L = 1 mL • 0.250 mol CuCl2 = 1 L • 2 mol Al reacts with 3 mol CuCl2 • 1 mol Al = 26.98 g Al • Plan: 200. mL CuCl2 L CuCl2  mol CuCl2  mol Al  g Al 200. mL CuCl2 = 0.899 g Al

  10. Solution Stoichiometry • What volume of 1.25 M NaOH is required to react with 0.800 L of 1.90 M H2SO4? • 2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l) • Given: 0.800 L H2SO4 • Want: ??? L NaOH • Conversion factors: • 1.90 mol H2SO4 = 1 L H2SO4 • 2 mol NaOH reacts with 1 mol H2SO4 • 1.25 mol NaOH = 1 L NaOH • Plan: 0.800 L H2SO4 mol H2SO4  mol NaOH  L NaOH 0.800 L H2SO4 = 2.43 L NaOH

  11. Titrations • A titration is a quantitative experiment designed to determine the unknown concentration of a solution by reacting it with another solution of known concentration. • Titrant – the solution with known concentration. • Analyte – the solution with unknown concentration. • Chemists add the titrant solution to the analyte solution until an equivalence point is reached. • Equivalence point – the point where the amount of titrant added is stoichiometrically equal to the amount of analyte present. • Depending on the reaction, there may be more than one equivalence point. • Chemists use indicators or meters to determine when the equivalence point has been reached.

  12. Titrations • 24.5 mL of an HCl solution with unknown concentration is titrated with 30.2 mL of 0.105 M NaOH solution. What is the molar concentration of the HCl solution? • HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) • Molarity = mol solute/L solution • We already know the volume of the solution: 24.5 mL • We can convert 24.5 mL  L • Molarity = mol solute / 0.0245 L solution 24.5 mL HCl = 0.0245 L HCl

  13. Titrations • Now we have to determine how many moles of HCl were in the original analyte sample. • Given: 30.2 mL of NaOH solution used • Want: ??? mol HCl in original sample • Conversion factors: • 0.001 L NaOH = 1 L NaOH • 0.105 mol NaOH = 1 L NaOH • 1 mol NaOH reacts with 1 mol HCl • Plan: 30.2 mL NaOH  L NaOH  mol NaOH  mol HCl 30.2 mL NaOH = 0.00317 mol HCl

  14. Titrations • The molar concentration of the HCl solution is: • M = 0.00317 mol HCl / 0.0245 L solution = 0.129 M

More Related