Stoichiometry Foldable - PowerPoint PPT Presentation

stoichiometry foldable n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Stoichiometry Foldable PowerPoint Presentation
Download Presentation
Stoichiometry Foldable

play fullscreen
1 / 11
Stoichiometry Foldable
1170 Views
Download Presentation
aziza
Download Presentation

Stoichiometry Foldable

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Stoichiometry Foldable

  2. Construction • Obtain 4 half sheets of paper • Fold three of the sheets into a foldable and staple the final sheet to the back

  3. Construction • Label the Tabs: • Stoichiometry • Mole Calculations • Mole Mole ratios • Volume Volume • Limiting Reactant • Excess Reagent • % Yield

  4. Mole Calculations • How many moles are in 2.5 g of nitrogen? You will use the Molar mass to solve this nitrogen is a HOFBrINCl element so you have to have two of them • 1 N= 14.01 g/mole N2= 28.02 g/1 mole • 2.5 g N2 x 1 mole =7.5 moles of N2 • 28.02 g • How many grams are in 3.5 moles of Nitrogen? • Once again you must use the molar mass • 3.5 moles N2 x 28.02 g =98.07 g = 98 g ( don’t forget sig.figs) • 1 mole

  5. Mole Mole • The coefficients in a balanced chemical equation are the number of moles. • In the equation: • N2 + 3H22 NH3 • In this reactions you need one mole of nitrogen for every three moles of Hydrogen and you will produce 2 moles of ammonia (NH3). This is the stoichiometric ratio 1N2 : 3 H2, 3 H2: 2NH3, 1 N2: 2 NH3 • But if you had 2.5 moles of nitrogen how much Hydrogen would you need to use the mole ratio of hydrogen to nitrogen • 2.5 moles of N2 x 3 H2 = 7.5 moles of H2 • 1 N2

  6. Volume  Volume • Avogadro’s Law- if everything in an equation is a gas the coefficients can also be volume • Given: N2 (g) + 3H2 (g)2NH3 (g) • If we had 3 Liters of N2 how much H2 would we need for complete reaction? • We will use the stoichiometric ratio again • 3 L N2 x 3 H2 = 9 L of H2 • 1 N2

  7. Limiting Reactant • The reactant that you will run out of first. • Given 25 g of N2 and 50 g of H2 what is your limiting reactant and how much product can you make? • N2 + 3H22NH3 • Step 1: Convert grams to moles ( this is what you have) • 25 g N2 x 1 mole =0.89 mole N2 • 28.02 g • 50 g H2 x 1 mole = 24.75 = 25 moles • 2.02 g

  8. Limiting Reactant • Step 2: Compare what you need with what you have • 0.89 moles N2 x 3 H2 = 2.67 moles of H2 • 1 N2 • 2.67 moles of H2 are needed to fully react with 0.89 moles of Nitrogen, since we have 25 moles we have plenty and will run out of the nitrogen first- therefore nitrogen is the limiting reactant

  9. Limiting reactant part 2 • You can only make as much product as your limiting reactant allows so once you know what your limiting reactant is convert moles of limiting to the moles of product. • 0.89 mole N2 x 2NH3 = 1.8 moles of NH3 will be produced • 1 N2 • 1.8 Moles NH3 x 17.04 g =30.67 = 31 g of NH3 • 1 mole

  10. Excess Reagent • In the previous example we were given 25 moles of Hydrogen but we only needed 2.67 moles of Hydrogen to fully react with the limiting reactant. The difference between what we have and what we used is the excess • 25 moles- 2.67 moles = 22.33 moles of Hydrogen • Converting the moles of excess to grams of excess • 22.33 moles x 2.02 g = 45 g H2 are in excess • 1 moles

  11. Percent Yield • Percent yield takes into account differences in experimental results and what is the calculated or theoretical results. The theoretical results are what is calculated based on the limiting reactant. • Example: In the lab 28 grams of NH3 are produced but the theoretical yield was 31 g • % yield = actual / theoretical X 100% • 28/31 = 90 %