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Stoichiometry: Mole-Mole Relationships and Mass Calculations

Learn how to use balanced chemical equations to predict the number of moles of products from a given amount of reactants and calculate the masses of reactants and products involved in a chemical reaction.

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Stoichiometry: Mole-Mole Relationships and Mass Calculations

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  1. Unit 6 Stochiometry

  2. Information Given by Chemical Equations • Example: CO(g) + 2H2(g) CH3OH(l) • The numbers we put in front of each element is called the coefficient. • The coefficient enables us to predict how much product we can get from a given quantity ofreactants.

  3. Let’s look at the synthesis of Methanol: • CO(g) + 2H2(g) CH3OH(l) • The coefficients give the relative number of molecules. We could multiply this equation by any number and still have a balanced equation. • 12 (CO(g) + 2H2(g) CH3OH(l)) • 12CO(g) + 24H2(g)  12CH3OH(l) • This equation is still a balanced equation. Since 12 represents a dozen, we could rewrite the equation in terms of dozens: • 1 dozen CO(g) + 2 dozen H2(g) 1 dozen CH3OH(l)

  4. Just as we multiplied the equation by 12, we could also multiply it by a very large number, say 6.022 x 1023. • 6.022x1023(CO(g) + 2H2(g) CH3OH(l)) • 6.022x1023CO(g) + 2(6.022x1023)H2(g)  6.022x1023CH3OH(l) • Just as we said 12 is a dozen, 6.022x1023 is 1 mole, so we could rewrite the equation in terms of moles: • 1 mol CO(g) + 2 mol H2(g) 1 mol CH3OH(l)

  5. Mole-Mole Relationships • Now that we known how to use moles in balanced equations we can use this information to predict the number of moles of products that a given number of reactants will yield. • Example: Consider the decomposition of water. • 2H2O(l) 2H2(g) + O2(g) • This equation tells us that 2 moles of H2O yields 2 moles of H2 and 1 mole of O2.

  6. Suppose we have 4 moles of water. If we decompose 4 mol of water, how many moles of products do we get? • One way to find the answer is to multiply the entire equation by 2, which gives us 4 mols of water. • 2(2H2O(l) 2H2(g) + O2(g)) • 4 H2O(l)  4H2(g) + 2O2(g) • Which could be written as:4 mol H2O(l)  4 mol H2(g) + 2 mol O2(g)

  7. Suppose we decompose 5.8 mols of water. How many mols of products are formed? 2H2O(l) 2H2(g) + O2(g) • Now it is not so easy to just multiply so we can use a mole ratio to calculate the amount of products. • A mole ratio is the ratio between two parts of the equation. • From the equation above we can see that 2 moles of H2O(l) produces 1 mole of O2 • We can write the above a as a ratio: 1mol O2 or like this 2mol H2O 2 mol H2O 1 mol O2

  8. We can use this mole ratio to calculate the O2 produced from 5.8 mol of H2O. • 5.8 mol H2O x 1 mol O2 = 2.9 mol O2 1 2 mol H2O • What would the mole ratio be between H2O and H2? • How many moles of H2 are produced with 5.8 moles of H2O? • 5.8 mol H2O x 1 mol H2 = 5. 8 mol H2 1 1 mol H2O

  9. Practice • Calculate the number of moles of oxygen required to react exactly with 4.30 moles of propane, C3H8, in the reaction described by the following unbalanced equation: C3H8(g) + O2(g) CO2(g) + H2O(g) • What is the mole ratio? • How many moles? • Calculate the number of moles of CO2 formed when 4.30 mol of C3H8 react with the O2. • Calculate the number of moles of NH3 that can be made from 1.30 moles of H2 reacting with excess N2. N2(g) + H2(g) NH3(g)

  10. Mole-Mole Relationships • Calculate the number of moles each product produced when 0.125 mol of Iron (II) Oxide reacts with Carbon. • FeO + C Fe + CO2 • Step 1: Balance the equation • Step 2: Use dimensional analysis to convert mols of FeO to mols of Fe, and mols of FeO to mols of CO2

  11. Mass Calculations • C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) • What type of reaction is this? • What mass of oxygen will be required to react exactly 44.1 g of C3H8?

  12. What we know: • The balanced equation for the reaction: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) • The mass of propane available : 44.1 g • What we want to calculate: • The mass of oxygen required to react exactly with all of the propane. • In a flow chart form we would write: 44.1 grams C3H8 ? grams of O2

  13. Overall plan to solve this problem: • We are given the number of grams of propane, so we must convert to moles of propane, because the balanced equation deals in moles rather than grams. • Next, we can use the coefficients in the balanced equation to determine the moles of oxygen required. • Finally, we will use the molar mass of O2 to calculate grams of oxygen. g C3H8 mol C3H8 mol O2 g O2

  14. Here is our balance equation: C3H8(g) + 5 O2(g)  3CO2(g) + 4H2O(g) 44.1 g C3H8 ? g of O2 ? mols C3H8 ? mols of O2 • What is the first thing we should do? • (Convert 44.1 g of C3H8 to moles)

  15. So 160. g of O2 is required to burn 44.1 g of propane. We could write this calculation as one full step. 44.1 gC3H8 x 1molC3H8 x 5mol O2 x 32.0gO2 1 44.09 gC3H8 1mol C3H8 1molO2

  16. Practice C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) • What mass of carbon dioxide is produced when 44.1 g of propane reacts with sufficient oxygen? • Calculate the mass of I2 needed to react with 35.0 g of Al. Al(s) + I2(s) Al I3(s)

  17. Steps for Calculating Masses of Reactants and Products • Balance the equation. • Convert the mass(g) to moles. • Use the balanced equation to set up the appropriate mole ratio(s). • Convert moles to moles. • Convert from moles back to mass(g).

  18. Practice • Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment. The products are solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide can 1.00x103 g of lithium hydroxide absorb?

  19. If chlorine gas is bubbled through a potassium iodide solution, elemental iodine is produce. • Calculate the mass of iodine produced when 4.50 x 103 g of chlorine gas is bubbled through an excess of potassium iodide solution.

  20. In the lab, you performed the reaction of a solid iron nail with copper (II) chloride producing copper solid and iron (III) chloride. Use the data from the data table to answers the questions below. • Mass of nail before reaction = 2.19g • Mass of nail after reaction = 1.67g • Mass of copper after reaction = 0.59 • Write a balanced chemical equation for this reaction. • Calculate the mass of iron used in the reaction. • Calculate the number of moles of iron used in the reaction. • Calculate the number of moles of copper that could be produced. • Calculate the mass of copper that could be produced?

  21. Percent Yield • Theoretical yield – the amount of product calculated to be produced in a reaction. • Actual yield – the amount of product you actually produced during a reaction.

  22. Calculating Percent Yield • Percent yield is used as a tool to calculate the comparison between the actual yield and theoretical yield. The closer to 100% the percent yield is, the better the results of the reaction. • Actual yield x 100 = Percent Yield Theoretical yield

  23. Practice • Methanol can be produced by the reaction between carbon monoxide and hydrogen. Suppose 6.85x104 g of CO is reacted with H2. The unbalanced reaction is: H2(g) + CO(g) CH3OH(l) a. Calculate the theoretical yield of methanol. b. If 3.57x104 g CH3OH is actually produced, what is the percent yield of methanol?

  24. The concept of Limiting Reactants • A restaurant prepares carryout lunch boxes. Each box consists of 1 sandwich, 3 cookies, 2 paper napkins, 1 milk carton, and 1 container. • The current inventory is 60 sandwiches, 102 cookies, 58 napkins, 41 milk cartons, and 66 cardboard containers. • As carryout lunch boxes are prepared, which item will be used up first? • How many lunch boxes can be made from each ingredient or reactant? • How many complete carryout lunch boxes can be assembled from this inventory?

  25. The concept of Limiting Reactants • If we were to make a s’more we would use the equations below: 1 chocolate + 3 marshmallows + 2 graham crackers  1 s’more • Determine all the mole ratios (or s’mole ratios) you can for this reaction. • Now, make as many s’mores as you can according to the reaction above given: • 5 ch + 21 mm + 12 gc • 9 ch + 18 mm + 8 gc • 4 ch. + 24 mm + 10 gc What is limiting the outcome? How much is left over?

  26. What if we had 20 moles of marshmallows, 7.3 moles of chocolate and 15 moles of graham crackers, how many s’mores could we make? • 20 mols mm x 1 mol sm = 6.7 mol sm 1 3 mol mm • 7.3 mol ch x 1 mol sm = 7.3 mol sm 1 1 mol ch • 15 mol gc x 1 mol sm = 7.5 mol sm 1 2 mol gc What is the smallest amount produced? 6.7 moles of product when 20 moles of marshmallows Which reactant limits how much is produced?

  27. We know that the marshmallows limit how much product we get, that makes the marshmallows the limiting reactant. • Now we wonder about what is left and how much? • So we have leftover chocolate and graham crackers, but how much is left. • If we know the amount of product produced, 6.7 moles, we can work backwards to calculate the amount of chocolate that was used in the reaction then subtract this from the starting amount and find out the amount left.

  28. So let’s first start with chocolate:We know the product is 6.7 moles of s’mores. 6.7 mols sm x 1 mol ch = 6.7 mol ch used 1 1 mol sm • 6.7 mol chocolate are used from the original 7.3 mol of chocolate. How much is unused? 7.3 mol – 6.7 mol = 0.6 mol chocolate unused • Calculate the mols of graham crackers leftover when 6.7 mols of s’mores are produced and there were 15 mols of graham crackers to start.

  29. Limiting Reactants • The reactant that runs out first and thus limits the amounts of products that can form is called the limiting reactant. • Let’s look closely at the steps involved in solving a limiting reactant problem.

  30. Steps for Solving Limiting Reactant Problems: • Write the balanced equation for the reaction. • Convert known masses of reactants to moles. (If you are in mols, do not do step 2!) 3. Using the numbers of moles of reactants and the appropriate mole ratios, compute which reactant is limiting by converting mols of reactants to mols of the same product. (The reactant that produces the smallestmols of product is the limiting reactant.) 4. Convert from moles of product to grams of product, using the molar mass. (If you are asked for mass of product)

  31. So let’s use these steps to solve a problem. • Suppose 25.0 kg (2.50x104 g) of nitrogen gas and 5.00 kg (5.00x103 g) of hydrogen gas are mixed and reacted to form ammonia. Calculate the mass of ammonia produced when this reaction is run to completion. • What is the balanced equation for this reaction? N2(g) + 3H2(g) 2NH3(g)

  32. Convert masses of reactants to moles. 2.50x104g N2 x 1 mol N2 = 8.92x102 mol N2 1 28.02 g N2 5.00x102g H2 x 1 mol H2 = 2.48x103 mol H2 1 2.016 g H2 • Use the moles of reactants to calculate the moles of products. 8.92x102 mol N2 x 2 mol NH3 = 1 1 mol N2 2.48x103 mol H2 x 2 mol NH3 = 1 3 mol H2

  33. Determine which is smaller and that is the limiting reactant. 1.784x103 mol NH3 produced from 8.92x102 mol N2 1.653x103 mol NH3 produced from 2.48x103 mol H2 • Calculate the mass of the product produced using mols of limiting reactant. 1.653x103 mol NH3 x 17.03g NH3 =2.81x104 g NH3 1 1 mol NH3

  34. Practice • If 3.6 grams of zinc react with 15 grams of HCl, what is the mass of ZnCl2 that is produced? What is the limiting reagent? What substance is left and how much? Zn(s) + HCl(aq) ZnCl2(aq) + H2(g)

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