Is it still balanced? Has anything changed? What does it mean? Is it still balanced? Has anything changed? What does it mean?

Transcription

1 Chemistry 11 Unit VII Stoichiometry Introduction: Let s look at the following equation 2 H 2 + O 2 2 H 2 O The numbers in front of the H 2 and H 2 O are called coefficients. What do they tell us? They tell us that 2 hydrogen molecules react with 1 oxygen molecule to produce 1 molecule of water. What about this equation 200 H O H 2 O Is it still balanced? Has anything changed? What does it mean? Yes it is still balanced, no nothing has changed and it means 200 hydrogen molecules combine with 100 oxygen molecules to produce 200 water molecules. What about this equation 2 x (6.02 x ) H x (6.02 x ) O 2 2 x (6.02 x ) H 2 O Is it still balanced? Has anything changed? What does it mean? Yes it is still balanced and no nothing has changed. To see what the equation means, let me write it a different way x = 1 mol, the above equation becomes 2 mol H mol O 2 2 mol H 2 O or simply 2 H 2 + O 2 2H 2 O Thus the coefficients in a balanced equation simply give us the ratio of reactants and products in terms of atoms and molecule or in terms of moles of atoms and molecules. Simmer Down Inc. 1

2 Stoichiometry is the relationship between the amount of reactants used in a chemical reaction and the amounts of products produced. Let s try a few examples to get your noodle fired up. Examples: Consider the following equation: N H 2 2 NH 3 1. How many molecules of Nitrogen are required to react with 27 molecules of Hydrogen? N H 2 2 NH 3 The balanced equation shows that 1 molecule of N 2 reacts with 3 molecules of H 2. To solve this problem just treat it like a unit conversion. 27 mlcs H 2 x 1 N 2 mlc = 9 N 2 mlcs 3 H 2 mlcs 2. How many molecules of NH 3 are produced if you have 10 molecules of H 2 (assume an excess of N 2 ) N H 2 2 NH 3 Excess is a very important concept in stoichiometry and one we will deal with a bit later. But for now, excess means that you have more than you need. In terms of stoichiometry, it means that all of the other reactant will be fully used up. In this question, it means that all 10 molecules of H 2 will react. 12 mlcs H 2 x 2 mlc NH 3 = 8 mlcs NH 3 3 mlcs H 2 3. How many moles of nitrogen are needed to react with 1.65 mol of hydrogen? N H 2 2 NH 3 4. How many moles of NH 3 are produced when 8.75 mol of N 2 is reacted with an excess of hydrogen? N H 2 2 NH 3 Simmer Down Inc. 2

3 Stoichiometry + moles + molecules + gas volume + mass = death What you need to realize is that in order to relate reactants to products, reactants to reactants etc, you HAVE TO BE IN MOLES. Once you re in moles you can use the balanced chemical equation and the Mole Highway (coefficients / mol ratios) to relate reactants and products. So remember this, if you are not given moles, convert into moles right away. The picture below may help. MASS of CHEMICAL #1 MASS of CHEMICAL #2 MOLECULES of CHEMICAL #1 MOLES of CHEMICAL #1 MOLE HWY MOLES of CHEMICAL #2 MOLECULES of CHEMICAL #2 VOLUME of CHEMICAL #1 VOLUME of CHEMICAL #2 Let s try some examples 2 C 2 H 6(g) + 7 O 2(g) 4 CO 2(g) + 6 H 2 O (l) 1. How many grams of H 2 O are produced if you react g of O 2 with an excess of C 2 H 6. Alrighty jump in! Starting with grams and ending with grams. We need to first get into moles of O 2 jump on the mole hwy and cross the mole bridge to mols of H 2 O and then back into grams easy O 2 x 1 mol O 2 x 6 mol H 2 O x 18.0 g H 2 O = = 22.0 g H 2 O 32.0 g 7 O 2 1 mol H 2 O Now it s time to hop in the mol car and get on the mol Hwy!! This is the stoichiometry step!!! ml of O 2, at S.T.P, was reacted with an excess of C 2 H 6. How many grams of carbon dioxide were produced? 2 C 2 H 6 (g) + 7 O 2 (g) 4 CO 2 (g) + 6 H 2 O (l) OK, same thing as above. In this case, we start with a volume at STP and end with grams. Again, convert to moles then back to grams. What your units!!! ml O 2 x 1 L x 1 mol O 2 x 4 mols CO 2 x 44.0 g CO 2 = = g CO ml 22.4 L 7 mols O 2 1 mol CO 2 Now it s time to hop in the mol car and get on the mol Hwy!! This is the stoichiometry step!!! Simmer Down Inc. 3

4 3. An ethane (C 2 H 6 ) burner is used in an auditorium as part of a chemistry demonstration. What volume of O 2 (g) at STP is consumed from the auditorium air if the burner produces 10.0 L of CO 2 (g) at STP during the demonstration? 2 C 2 H 6 (g) + 7 O 2 (g) 4 CO 2 (g) + 6 H 2 O (l) 4. A sample of porous, gas-bearing rock is crushed and 1.35 x 10-6 g of C 3 H 8 (g) is extracted from the powdered rock. How many molecules of CO 2 are produced if the gas sample is burned in the presence of an excess of O 2 (g)? C 3 H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H 2 O (l) 5. How many O 2 atoms are required to produce 7.98 x molecules of CO 2? (Assume an excess of ethane (C 2 H 6 ) ) 2 C 2 H 6 (g) + 7 O 2 (g) 4 CO 2 (g) + 6 H 2 O (l) Simmer Down Inc. 4

5 Stoichiometry + Molarity = More fun More of the same. Make sure you convert to moles first. Remember Molarity = Moles or Moles = Molarity x Volume (L) Volume (L) Let s try some more fab-u-lous-ly fun examples. Examples: CaCO 3(s) + 2 HCl (aq) CaCl 2 (aq) + CO 2 (g) + H 2 O (l) 1. A tablet of Tums (mostly CaCO 3 ) has a mass of g. What volume of stomach acid having [HCl] = M is neutralized by a g portion of CaCO 3. Two ways to solve this one. 1 st Method Two Step (not the country dance people thank goodness ) moles of CaCO 3 = g CaCO 3 x 1 mol g = moles CaCO 3 moles of HCl = moles CaCO 3 x 2 HCl 1 CaCO 3 = moles HCl Now it s time to hop in the mol car and get on the mol Hwy!! This is the stoichiometry step!!! Volume of HCl = n V = mol mol/l = 15 L 2 nd Method Unit Conversions This is molarity as a unit conversions Vol of HCl = g CaCO 3 x 1 mol g x 2 HCl x 1 CaCO 3 1 L mol = 15 L Now it s time to hop in the mol car and get on the mol Hwy!! This is the stoichiometry step!!! Simmer Down Inc. 5

6 2. What volume of CO 2 (g) at STP is produced if 1.85 L of M HCl reacts with an excess of CaCO 3? CaCO 3(s) + 2 HCl (aq) CaCl 2 (aq) + CO 2 (g) + H 2 O (l) Need to find moles of HCl so we can use the mol Hwy. Again, you can use two different methods. Method #1 Two Step mole of HCl = mol L x 1.25 L = mol HCl (don t round off!!!) moles of CO moles HCl x 1 CO 2 2 HCl = mol CO 2 volume of CO mol x 22.4 L 1 mol = L Method #2 Unit conversion volume of CO mol L x 1.25 L x 1 CO 2 2 HCl x 22.4 L 1 mol = L 3. How many molecules of water are produced when ml of M HCl is reacted with an excess of CaCO 3? CaCO 3(s) + 2 HCl (aq) CaCl 2 (aq) + CO 2 (g) + H 2 O (l) Simmer Down Inc. 6

7 Limiting Reagents / Excess Reactants So far we have assumed that a given reactant is completely used up during the reaction. In reality, reactions are often carried out in such a way that one or more of the reactants are present in EXCESS amounts. Some reasons for having an excess amount include: (i) Adding an excess of one reactant to make sure all of a second reactant is completely used (the second reactant may be too expensive to waste or harmful to the environment). (ii) Unavoidably having a reactant in excess because a limited amount of another reactant is available. In these types of calculations, the reactant that is present in lesser amount is called the Limiting reactant. Since the limiting reactant gets completely used up first, it sets the limit on the amount of product that can be formed and the amount of the excess reactant used in the reaction. LIMITING REACTANT: is the reactant that sets a limit on the amount of product that can be formed (completely used up in a chemical reaction). EXCESS REACTANT: is the reactant that is not completely used up in a chemical reaction. *The limiting reactant determines the yield of the product (how much product(s) will form) A simple analogy Imagine you work at McDonalds. You have 13 hamburger buns and 9 beef patties. How many regular hamburgers can you make? You could make 9 hamburgers. What / how much was left over? There would be 4 hamburger buns in EXCESS!! Therefore, the beef patty is known as the LIMITING ingredient since it limits or determines how many regular hamburgers can be made. ***Note: We do not predict based on the number of hamburger buns, but rather the patties as they are LIMITING. Simmer Down Inc. 7

8 Here are some helpful hints for determining which reactant is limiting and which is in excess. 1. Make sure you have a BALANCED chemical equation. 2. Write down the amounts of each reactant you start with. 3. ***MOST IMPORTANT STEP*** Choose one of the reactants. Starting with that reactant, use stoichiometry to figure out how much of the other reactant you need to completely react (use up). This is the amount you NEED to fully react. Once calculated, start with the other reactant and perform the same calculation. 4. Looking at either reactant, compare the amount you start with (amount you HAVE) with the amount you NEED (calculated value in step 2). If you HAVE more than you NEED, that reactant is in excess. If you HAVE less than you need, that reactant is the limiting reactant / reagent. 5. Once you have found the limiting reagent, use it to determine how much of the other reactant is used in the reaction and then you can calculate the amount left over (in excess) Examples 1. If 20.0 g of Hydrogen gas reacts with g of Oxygen, which reactant is present in excess and by how much? Balanced equation is: 2H 2 (g) + O 2 (g) 2H 2 O (l) Simmer Down Inc. 8

9 2. If 79.1 g of Zn reacts with 60.0 g HCl a. Which reactant is in excess and by how much? b. What is the mass of each product? Balanced equation: Zn + 2HCl ZnCl 2 + H g Na 2 CO 3 reacts with g of AgNO 3. a. Which reactant is in excess and by how much? b. Calculate how much solid Ag 2 CO 3 is produced Na 2 CO 3 (aq) + 2 AgNO 3 (aq) 2 NaNO 3 (aq) + Ag 2 CO 3(s) Simmer Down Inc. 9

10 Percent Yield: An assumption we have been making is that every reaction goes to completion. Completion means that ALL the limiting reactant / reagent has been converted into product, leaving only the excess reactant / reagent. Often 100% of the expected amount of product cannot be obtained from a reaction. The term Percent Yield is used to describe the amount of product actually obtained as a percentage of the expected amount. Reasons for reduced yields The reactant may not all react because o Not all of the pure material actually reacts o The reactants may be impure Some of the product is lost during procedures such as solvent extraction, filtration etc. The equation of percent yield is as follows % Yield = Actual Yield x 100% Theoretical Yield Actual Yield = amount of product obtained (determined experimentally). Theoretical Yield = amount of product expected (determined from calculations based on the Stoichiometry of the reaction). ***Amounts can be expressed in grams, moles, molecules *** The percent yield MUST be less than 100%. When calculating the theoretical yield ASSUME a 100% yield. Simmer Down Inc. 10

11 Examples: 1. When 15.0 g of CH 4 is reacted with an excess of Cl 2 according to the reaction: CH 4 + Cl 2 CH 3 Cl + HCl A total of 29.7 g of CH 3 Cl is formed. Calculate the percent yield. The actual yield is given to use (29.7 g); therefore, we need to calculate the theoretical yield. Start with what you are given g CH 4 x 1 mol CH 4 x 1 mol CH 3 Cl x 50.5 g CH 3 Cl = 47.3 g CH 3 Cl 16.0 g CH 4 1 mol CH 4 1 mol CH 3 Cl % Yield = Actual Yield x 100% Theoretical Yield = 29.7 g CH 3 Cl x 100 = 62.8 % 47.3 g CH 3 Cl g of K 2 CO 3 is produced when g of KO 2 is reacted with an excess of CO 2? The reaction is: 4KO 2(s) + 2CO 2 (g) 2K 2 CO 3(s) + 3O 2 (g) Simmer Down Inc. 11

12 3. What mass of K 2 CO 3 is produced when 1.50 g of KO 2 is reacted with an excess of CO 2 if the reaction has a 76% yield? The reaction is: 4KO 2(s) + 2CO 2 (g) 2K 2 CO 3(s) + 3O 2 (g) Simmer Down Inc. 12