Presentation on theme: "Stoichiometry – “Fun With Ratios”"— Presentation transcript:
1Stoichiometry – “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
2Ex. 2H2 + O2 2H2O Interpretations: 2 molecules of H2 combine with 1 molecule of O2 to produce 2 molecules of H2O.Or2 moles of H2 combine with 1 mole of O2 to produce 2 moles of H2O.
3As long as we are dealing with amounts of molecules, the ratios will stay the same: 2 H2 : 1 O2: 2 H2O6 moles H2 ? Moles O2Use Mole Ratio 1 mole O22 moles H2
4Every Stoich Problem uses Mole Ratios Step 1-Write the balanced equationIf 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.
5Step 3- Solve problem using unit factors. Units are your friends!!Step 4- Check significant digitsOh no, not sig digs again!
6Molex 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 > MgOEquation 2Mg + O2 > 2 MgOWork: Moles O Moles of MgO /1 Mol O2Answer: 3.86 moles MgO
7Molex Gramy, Gramx Moley Ex. How many grams of oxygen will react with 4.25 moles of magnesium?Work: Balanced Equation 2Mg + O2 > 2 MgO4.25 Moles Mg 1 Mole O2/2 Moles Mg g/1 mol O2Answer: 68.0 g O2Ex. How many moles of water are produced when 42.0 g of methane, CH4, burns in the presence of oxygen gas?Work: Balanced Equation CH O2 > CO2 + 2H2O42.0 g of CH mole CH4/ g 2 moles H2O/1 Mole CH4Answer: 5.25 mol H2O
8Gramx Gramy This is how we use stoichiometry in the REAL WORLD. There are four steps in solving gramgram1. Write correct balanced equation2. Convert grams given into moles3. Use mole ratio based on ratio between the given and the unknown using coefficients from balanced equation.4. Convert moles of unknown into grams.
10Ex. How many grams of AlCl3 can be made by reacting 100 Ex. How many grams of AlCl3 can be made by reacting grams of chlorine gas with excess aluminum metal?3Cl2(g) + 2Al(s) > 2AlCl3Answer: g AlCl3
11How many grams of NaOH is produced from 1.20 x 103 g of Na2O? Ex. Given the equation:Na2O + H2O NaOHHow many grams of NaOH is produced from 1.20 x 103 g of Na2O?Work: Balance Na2O + H2O 2NaOH1.20 x 103 g Na2O 1 mol Na2O mol NaOH/ g/ g 1 mol Na2O 1 mol NaOHAnswer: 1.55 x 103 g NaOH
12Limiting 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.
13Practice ProblemsConsider the reaction:2 Al I2 2 AlI3Determine 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 iodineAlI2
14B. I2 Which is limiting Reactant? A. Al 1.2 moles of Al moles AlI moles AlI moles Al2.40 moles I moles AlI moles AlI23 moles I2Al is limiting reactant
15b). 1.20 grams of Al and 2.40 grams of iodine A 1.20g AlB g I21.20 g Al 1mol Al mol AlI mols AlI2g 2 mol Al2.40 g I2 1 mol I mol AlI mols AlI2g 3 mole I2
16How many grams of excess reactant will remain? .006 mols AlI moles Al g2mole AlI mole Al.0179 g g excess
1715.00 g aluminum sulfide and mL water react until the limiting reagent is used up. Here is the balanced equation for the reaction.Al2S H2O 2 Al(OH) H2Sa.) 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?
18Percent 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.
19Example: In the reaction between nitrogen gas and hydrogen gas producing ammonia: a.) If you 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?
202 Al + 3 CuCl2 2 AlCl3 + 3 Cu How many grams of aluminum will be required to completely replace copperfrom a 208 mL of a 0.100M solution ofcopper(II) chloride?2 Al CuCl2 2 AlCl CuAnswer = g Al