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Stoichiometry – “Fun With Ratios”

Stoichiometry – “Fun With Ratios”. Main Idea: The coefficients from the balanced equation tell the ratios between reactants and products. This ratio applies to #’s of molecules (or moles) not masses . It’s all about the moles. Ex. 2 H 2 + O 2  2 H 2 O. Interpretations:

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Stoichiometry – “Fun With Ratios”

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  1. Stoichiometry – “Fun With Ratios” • Main Idea: The coefficients from the balanced equation tell the ratios between reactants and products. • This ratio applies to #’s of molecules (or moles) not masses. • It’s all about the moles

  2. Ex. 2H2 + O2 2H2O • Interpretations: 2 molecules of H2 combine with 1 molecule of O2 to produce 2 molecules of H2O. Or 2 moles of H2 combine with 1 mole of O2 to produce 2 moles of H2O.

  3. As long as we are dealing with amounts of molecules, the ratios will stay the same: 2 H2 : 1 O2: 2 H2O 6 moles H2 ? Moles O2 Use Mole Ratio  1 mole O2 2 moles H2

  4. Every Stoich Problem uses Mole Ratios • Step 1-Write the balanced equation If you haven’t written the correct equation & haven’t balanced it right, you will not solve the problem correctly. • Step 2-Draw a road map for solving the problem. At the very least you will use the mole ratio to determine the moles of what you are trying to find. If given grams get the moles 1st. If your answer should be in grams, convert from moles to grams.

  5. Step 3- Solve problem using unit factors. Units are your friends!! • Step 4- Check significant digits Oh no, not sig digs again!

  6. Molex Moley Conversions • Ex. How many moles of magnesium oxide will be produced if 1.93 moles of O2 reacts with magnesium completely. • Road Map Moles O2 > MgO • Equation 2Mg + O2 > 2 MgO Work: 1.93 Moles O2 2 Moles of MgO /1 Mol O2 Answer: 3.86 moles MgO

  7. Molex Gramy, Gramx  Moley • Ex. How many grams of oxygen will react with 4.25 moles of magnesium? Work: Balanced Equation 2Mg + O2 > 2 MgO 4.25 Moles Mg 1 Mole O2/2 Moles Mg 31.98 g/1 mol O2 Answer: 68.0 g O2 • Ex. How many moles of water are produced when 42.0 g of methane, CH4, burns in the presence of oxygen gas? • Work: Balanced Equation CH4 + 2O2 > CO2 + 2H2O • 42.0 g of CH4 1 mole CH4/ 16.0 g 2 moles H2O/1 Mole CH4 Answer: 5.25 mol H2O

  8. Gramx Gramy • This is how we use stoichiometry in the REAL WORLD. There are four steps in solving gramgram 1. Write correct balanced equation 2. Convert grams given into moles 3. Use mole ratio based on ratio between the given and the unknown using coefficients from balanced equation. 4. Convert moles of unknown into grams.

  9. Ex. How many grams of AlCl3 can be made by reacting 100.0 grams of chlorine gas with excess aluminum metal? 3Cl2(g) + 2Al(s) > 2AlCl3 Answer: 125.3 g AlCl3

  10. Ex. Given the equation: Na2O + H2O  NaOH How many grams of NaOH is produced from 1.20 x 103 g of Na2O? Work: Balance Na2O + H2O  2NaOH • 1.20 x 103 g Na2O 1 mol Na2O2 mol NaOH/ 40.0g/ 62.0g 1 mol Na2O 1 mol NaOH • Answer: 1.55 x 103 g NaOH

  11. Limiting Reactants and Excess • What is the Limiting Reagent (Reactant)? It is the substance in a chemical reaction that runs out first. • The limiting reactant (reagent) determines how much product you can make. • If you are given amounts of more than one reactant, determine how much product you can make with each of them. Whichever produces the LEAST amount of product is your limiting reagent.

  12. Practice Problems Consider the reaction: 2 Al + 3 I2 2 AlI3 Determine the limiting reagent and the theoretical yield of the product, aluminum iodide, if one starts with: a). 1.20 moles of Al and 2.40 moles of iodine • Al • I2

  13. Which is limiting Reactant? A. Al B. I2 • 1.2 moles of Al 2 moles AlI2 1.2 moles AlI2 2 moles Al • 2.40 moles I2 2 moles AlI2 1.6 moles AlI2 3 moles I2 • Al is limiting reactant

  14. b). 1.20 grams of Al and 2.40 grams of iodine A 1.20g Al B 2.40 g I2 1.20 g Al 1mol Al 2 mol AlI2 .0447 mols AlI2 26.9815 g 2 mol Al 2.40 g I2 1 mol I22 mol AlI2 .00630 mols AlI2 253.8088 g 3 mole I2

  15. How many grams of excess reactant will remain? .006 mols AlI2 2 moles Al26.9853g 2mole AlI2 1 mole Al .0179 g 2.38 g excess

  16. 15.00 g aluminum sulfide and 12.00 mL water react until the limiting reagent is used up. Here is the balanced equation for the reaction. Al2S3 + 6 H2O  2 Al(OH)3 + 3 H2S a.) What is the maximum mass of H2S which can be formed from these reagents ? b). Which is the limiting reagent? c). How much of the excess reactant is used up?

  17. Percent Yield Percent Yield = (Actual Yield / Theoretical Yield) * 100 • Theoretical Yield is the amount that your stoichiometric calculations have predicted. • Actual Yield is what you actually produced in an experiment.

  18. Example: In the reaction between nitrogen gas and hydrogen gas producing ammonia: a.) Ifyou start with 14 grams of N2 and 6.0 grams of H2, what will be the limiting reagent and the theoretical yield? b.) If 10. grams of the product was formed, what would be the percent yield?

  19. How many grams of aluminum will be required to completely replace copper from a 208 mL of a 0.100M solution of copper(II) chloride? 2 Al + 3 CuCl2 2 AlCl3 + 3 Cu Answer = 0.374 g Al

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