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Stoichiometry – “Fun With Ratios” Main Idea: The coefficients from the balanced equation tell the ratios between reactants and products. This ratio applies.

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Presentation on theme: "Stoichiometry – “Fun With Ratios” Main Idea: The coefficients from the balanced equation tell the ratios between reactants and products. This ratio applies."— Presentation transcript:

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. 2H 2 + O 2  2H 2 O Interpretations: 2 molecules of H 2 combine with 1 molecule of O 2 to produce 2 molecules of H 2 O. Or 2 moles of H 2 combine with 1 mole of O 2 to produce 2 moles of H 2 O.

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

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 1 st. 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 Mole x  Mole y Conversions Ex. How many moles of magnesium oxide will be produced if 1.93 moles of O 2 reacts with magnesium completely. Road Map Moles O 2 > MgO Equation 2Mg + O 2 > 2 MgO Work: 1.93 Moles O 2 2 Moles of MgO /1 Mol O 2 Answer: 3.86 moles MgO

7 Mole x  Gram y, Gram x  Mole y Ex. How many grams of oxygen will react with 4.25 moles of magnesium? Work: Balanced Equation 2Mg + O 2 > 2 MgO 4.25 Moles Mg 1 Mole O 2 /2 Moles Mg g/1 mol O 2 Answer: 68.0 g O 2 Ex. How many moles of water are produced when 42.0 g of methane, CH 4, burns in the presence of oxygen gas? Work: Balanced Equation CH 4 + 2O 2 > CO 2 + 2H 2 O 42.0 g of CH 4 1 mole CH 4/ 16.0 g 2 moles H 2 O/1 Mole CH 4 Answer: 5.25 mol H 2 O

8 Gram x  Gram y 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.

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10 Ex. How many grams of AlCl 3 can be made by reacting grams of chlorine gas with excess aluminum metal? 3Cl 2 (g) + 2Al(s) > 2AlCl 3 Answer: g AlCl 3

11 Ex. Given the equation: Na 2 O + H 2 O  NaOH How many grams of NaOH is produced from 1.20 x 10 3 g of Na 2 O? Work: Balance Na 2 O + H 2 O  2NaOH 1.20 x 10 3 g Na 2 O 1 mol Na 2 O 2 mol NaOH/ 40.0g/ 62.0g1 mol Na 2 O1 mol NaOH Answer: 1.55 x 10 3 g NaOH

12 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.

13 Practice Problems Consider the reaction: 2 Al + 3 I 2  2 AlI 3 Determine the limiting reagent and the theoretical yield of the product, aluminum iodide, if one starts with: a) moles of Al and 2.40 moles of iodine A. Al B. I 2

14 Which is limiting Reactant? A. Al B. I moles of Al 2 moles AlI moles AlI2 2 moles Al 2.40 moles I 2 2 moles AlI moles AlI 2 3 moles I 2 Al is limiting reactant

15 b) grams of Al and 2.40 grams of iodine A 1.20g Al B 2.40 g I g Al 1mol Al 2 mol AlI mols AlI g 2 mol Al 2.40 g I 2 1 mol I 2 2 mol AlI mols AlI g 3 mole I 2

16 How many grams of excess reactant will remain?.006 mols AlI 2 2 moles Al g 2mole AlI 2 1 mole Al.0179 g 2.38 g excess

17 15.00 g aluminum sulfide and mL water react until the limiting reagent is used up. Here is the balanced equation for the reaction. Al 2 S H 2 O  2 Al(OH) H 2 S a.) What is the maximum mass of H 2 S which can be formed from these reagents ? b). Which is the limiting reagent? c). How much of the excess reactant is used up?

18 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.

19 Example: In the reaction between nitrogen gas and hydrogen gas producing ammonia: a.) If you start with 14 grams of N 2 and 6.0 grams of H 2, 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?

20 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 CuCl 2  2 AlCl Cu Answer = g Al


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