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Chemical Equations and Reaction Stoichiometry. Chemical equations are used to describe chemical reactions, and they show (1)The substances that react,

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Presentation on theme: "Chemical Equations and Reaction Stoichiometry. Chemical equations are used to describe chemical reactions, and they show (1)The substances that react,"— Presentation transcript:

1 Chemical Equations and Reaction Stoichiometry

2 Chemical equations are used to describe chemical reactions, and they show (1)The substances that react, called reactants; (2)the substances formed, called products; and (3) the relative amounts of the substances involved. CHEMICAL EQUATIONS N.B. The amounts of substances are in moles Number of moles

3 What does this equation tell us? it says that for every CH 4 molecule that reacts, two molecules of O 2 also react, and that one CO 2 molecule and two H 2 O molecules are formed (produced). If we multiply throughout by Avogadro's number So, it turns out that The equation, therefore, tells us that for every CH 4 mole that reacts, two moles of O 2 also react, and that one CO 2 mole and two H 2 O moles are formed (produced).

4 We know the mass of 1 mol (molar mass) of each of these substances, Molar mass of CH 4 = 12.0 g mol x 1.0 g mol -1 = 16 g mol -1 Molar mass of O 2 = 2 x 16.0 g mol -1 = 32.0 g mol -1 Molar mass of CO 2 = 12.0 g mol x 16.0 g mol -1 = 44.0 g mol -1 Molar mass of H 2 O = 2 x 1.0 g mol g mol -1 = 18.0 g mol -1 so we can also write The equation then tells us that 16.0 g of CH 4 reacts with 64.0 g of O 2 to produce 44.0 g of CO 2 and 36.0 g of H 2 O

5 In fact knowing the BALANCED chemical equation allows us to know a great deal of information; we can calculate for example;  Number of Moles Formed/Produced  Mass of a Reactant Required  Mass of a Product Formed  The limiting reactant  Percent yield How?

6 How many moles of water (H 2 O) could be produced by the reaction of 3.5 mol of methane (CH 4 ) with excess oxygen (O 2 ) (i.e., more than a sufficient amount of oxygen is present)? 3.5 mol x MAKE SURE THAT THE EQUATION IS BALANCED; that is the number of atoms of each element that appears in the LHS appears as well in the RHS. If it is balanced then we can read the equation as 3.5 moles x Then 7.0 moles of water will be produced Number of Moles Formed

7 What mass of oxygen is required to react completely with 1.20 mol of CH 4 ? answer The balanced equation tells us but we are asked about the mass. We therefore need to calculate the mass of 2.40 moles of oxygen moles x Then 2.40 moles of oxygen are required. Mass of one mole of oxygen= molar mass of oxygen (O 2 ) =16.0 g mol -1 x 2= 32.0 g mol -1 The required mass of oxygen (O 2 )=2.40 mol x 32.0 g mol -1 =76.8 g Mass of a Reactant Required

8 What mass of oxygen is required to react completely with 24.0 g of CH 4 ? The molar mass of O 2 = 2 x 16.0 g mol -1 =32.0 g mol -1 The molar mass of CH 4 = 12.0 g mol x 1.0 g mol -1 = 16.0 g mol -1 The balanced equation tells us One molar mass of CH 4 (M CH4 ) reacts with 2 molar masses of O 2 (M O2 ) M CH4 2M O g X Then 96.0 g of oxygen are required Mass of a Reactant Required

9 Phosphorus, P 4, burns with excess oxygen to form tetraphosphorus decoxide, P 4 O 10. In this reaction, what mass of P 4 reacts with 1.50 moles of O 2 ?

10 Mass of a Product Formed Calculate the mass of CO 2, in grams, that can be produced by burning 6.00 mol of CH 4 in excess O 2. The molar mass of CO 2 = 12.0 g mol x 16.0 g mol -1 =44.0 g mol -1 The molar mass of CH 4 = 12.0 g mol x 1.0 g mol-1 = 16.0 g mol -1 The balanced equation tells us One molar mass of CH 4 produces ONE molar masses of CO molar mass of CH4 produces X

11 THE CONCEPT OF LIMITING REACTANT Say we want to me make a sandwich; we then need two slices of bread and one slice of cheese (The sandwich-making equation) What if we have six slices of bread and two slices of cheese. Even though you have extra bread, you are limited to making two sandwiches by the amount of cheese you have on hand. Cheese is the limiting reactant. So the number of sandwiches is determined by the ratio of cheese slices to those of bread

12 When an equal number of molecules of H 2 and O 2 are reacted according to the equation 2H 2 +O 2 → 2H 2 O all of the H 2 completely reacts, whereas only half of the O 2 is consumed. In this case, the H 2 is the limiting reactant and the O 2 is the excess reactant.

13 What is the limiting reactant in the reaction of 16.0 g of CH 4 with 48.0 g of O 2 according to the following equation? We have 16.0 g of CH 4 = 1.0 mol of CH 4 We have 48.0 g of O 2 = 1.5 mol of O 2 Which one is in excess? Do we have enough CH 4 to react with 1.5 mol of O 2 ? The balanced equation tells us 1 mol (CH 4 ) reacts with 2 mol (O 2 ) X1.5 mol (O 2 ) X=1.5 mol x 1.0 mol / 2.0 mol = 0.75 mol Then the number of moles of CH 4 required to react with 1.5 mol of O 2 = 0.75 mol We have 1.0 mol of CH 4 i.e. we have excess of CH 4 Do we have enough O 2 to react with 1.0 mol of CH 4 ? The balanced equation tells us 1 mol (CH 4 ) reacts with 2 mol (O 2 ) 1.0 mol (CH 4 ) X X=1.0 mol x 2.0 mol / 1.0 mol = 2.0 mol Then the number of moles of O 2 required to react with 1.0 mol of CH 4 = 2.0 mol we have only 1.5 mol of O 2 i.e. O 2 is the limiting reactant Then we should base any further calculation on the amount of O 2

14 What mass of CO 2 could be formed by the reaction of 16.0 g of CH 4 with 48.0 g of O 2 ? We have 16.0 g of CH 4 = 1.0 mol of CH 4 We have 48.0 g of O 2 = 1.5 mol of O 2 Which one is in excess? Do we have enough CH 4 to react with 1.5 mol of O 2 ? The balanced equation tells us 1 mol (CH 4 ) reacts with 2 mol (O 2 ) X1.5 mol (O 2 ) X=1.5 mol x 1.0 mol / 2.0 mol = 0.75 mol Then the number of moles of CH 4 required to react with 1.5 mol of O 2 = 0.75 mol We have 1.0 mol of CH 4 i.e. we have excess of CH 4 Do we have enough O 2 to react with 1.0 mol of CH 4 ? The balanced equation tells us 1 mol (CH 4 ) reacts with 2 mol (O 2 ) 1.0 mol (CH 4 ) X X=1.0 mol x 2.0 mol / 1.0 mol = 2.0 mol Then the number of moles of O 2 required to react with 1.0 mol of CH 4 = 2.0 mol we have only 1.5 mol of O 2 i.e. O 2 is the limiting reactant Then we should base any further calculation on the amount of O 2 In here, we are not told, as in previous examples, that one of the reactants exists in excess so we cannot be sure the amount of which reactant should be used in the calculation. So, first we need to know the limiting reactant

15 What mass of CO 2 could be formed by the reaction of 16.0 g of CH 4 with 48.0 g of O 2 ? (Continued) The balanced equation tells us 2 M O2 produce M CO g X We base our calculation on the amount of O 2 because it is the limiting reactant

16 What mass of CO 2 could be formed by the reaction of 16.0 g of CH 4 with g of O 2 ? The balanced equation tells us one molar mass of CH 4 produces ONE molar mass of CO g produces X We have 16.0 g of CH 4 = 1.0 mol of CH 4 We have g of O 2 = 4.0 mol of O 2 The balanced equation tells us that one mole of CH 4 reacts with two moles of O 2. Then the number of moles of CH 4 required to react with 4.0 mol of O 2 =2.0 mol we only have 1.0 mol of CH 4 i.e. we do not have excess of CH 4 The balanced equation tells us that one mole of CH 4 reacts with two moles of O 2. Then the number of moles of O 2 required to react with 1.0 mol of CH 4 = 2.0 mol we have 4.0 mol of O 2 i.e. we have excess of O 2 CH 4 is therefore the limiting reactant Then we should base our calculation on the amount of CH 4

17 What mass of CO 2 could be formed by the reaction of 32.0 g of CH 4 with g of O 2 ? We have 32.0 g of CH 4 = 2.0 mol of CH 4 We have g of O 2 = 4.0 mol of O 2 The balanced equation tells us that one mole of CH4 reacts with two moles of O 2. Then the number of moles of CH 4 required to react with 4.0 mol of O 2 = 2.0 mol we already have 2.0 mol of O 2 Then the number of moles of O 2 required to react with 2.0 mol of CH 4 = 4.0 mol we already have 4.0 mol of CH 4 Then the reactants exist in soichiometric ratio and therefore there is no limiting reactant. We can base our calculation on the amount of any one of them. The balanced equation tells us one molar mass of CH 4 produces ONE molar mass of CO g produces X

18 PERCENT YIELDS FROM CHEMICAL REACTIONS The theoretical yield from a chemical reaction is the yield calculated by assuming that the reaction goes to completion. In practice we often do not obtain as much product from a reaction mixture as is theoretically possible. This is true for several reasons. (1) Many reactions do not go to completion; that is, the reactants are not completely converted to products. (2) In some cases, a particular set of reactants undergoes two or more reactions simultaneously, forming undesired products (side reactions) as well as desired products. (3) In some cases, separation of the desired product from the reaction mixture is so difficult that not all of the product formed is successfully isolated. The term percent yield is used to indicate how much of a desired product is obtained from a reaction.

19 A 15.6 gram sample of C 6 H 6 was mixed with excess HNO 3. We isolated 18.0 grams of C 6 H 5 NO 2. What is the percent yield of C 6 H 5 NO 2 in this reaction? answer First we interpret the balanced chemical equation We do not need to work out which is the limiting reactant as we are told that we have excess of HNO 3 The balanced equation tells us One molar mass of C 6 H 6 produces ONE molar mass of C 6 H 5 NO g produces X

20 CONCENTRATIONS OF SOLUTIONS The reaction between two solid reactants often proceeds very slowly or not at all. Many chemical reactions are therefore more conveniently carried out with the reactants mixed in solution rather than as pure substances. A solution is a homogeneous mixture, at the molecular level, of two or more substances. Simple solutions usually consist of one substance, the solute, dissolved in another substance, the solvent. The solutions used in the laboratory are usually liquids, and the solvent is often water. These are called aqueous solutions. The general term concentration refers to the quantity of solute in a standard quantity of solution.

21 The unit of molarity is therefore mol/L, mol L -1 or M “read as molar ” Molar concentration, or molarity (M), is defined as the number of moles of solute dissolved in one litre (cubic decimeter) of solution. E.g. An aqueous solution that is 0.15 M NH 3 (read this as “0.15 molar NH 3 ”) contains 0.15 mol NH 3 per litre of solution. If you want to prepare a solution that is, for example, M CuSO4, you place mol CuSO4 in a L volumetric flask or place mol CuSO4 in a 0.5 L volumetric flask or …

22 A sample of NaNO 3 weighing 0.38 g is placed in a 50.0 mL volumetric flask. The flask is then filled with water to the mark on the neck, dissolving the solid. What is the molarity of the resulting solution? To calculate the molarity, we need the number of moles of solute. Therefore, you first convert grams NaNO 3 to moles. Molar mass of NaNO 3 = 23.0 g mol g mol x16.0 g mol -1 =84.9 g mol -1 read this as “0.089 molar”

23 How many grams of CuSO 4.5H 2 O do we need in order to prepare a 0.2 M solution of CuSO 4.5H 2 O in a 250 mL volumetric flask. The number of moles required is 0.05 mol of CuSO 4.5H 2 O But, we are asked about the mass ( how many grams?) We can calculate the mass of one mole of CuSO 4.5H 2 O i.e. its molar mass then we multiply this molar mass by the number of moles (0.1 mol) to get the required mass So we need to dissolve g of CuSO 4.5H 2 O in 250 mL water to make 0.2 M solution

24 mol CuSO 4. 5H 2 O (12.48 g) The solute is transferred carefully to the volumetric flask Water (solvent) is added to bring the solution level to the mark on the neck of the 250- mL volumetric flask. The molarity of the solution is mol/0.250 L= M. Let’s prepare a 0.2 M solution of CuSO 4. 5H 2 O

25 Calculate the mass of barium hydroxide Ba(OH) 2 required to prepare 2.50 L of a 0.06 M solution of barium hydroxide.

26 We need to add to a reaction vessel g of sodium hydroxide, NaOH, in aqueous solution. How many milliliters of M NaOH should be added?

27 DILUTING SOLUTIONS Number of moles before dilution=Number of moles after dilution=n Note, in here, M is the molarity of the solution (not the molar mass of the solute)

28 How many milliliters of 18.0 M H 2 SO 4 are required to prepare 1.00 L of a 0.900M solution of H 2 SO 4 ?

29 Arrange the beakers in order of increasing concentration

30 Solution Stoichiometry If we plan to carry out a reaction in a solution, we must calculate the amounts of solutions that we need. If we know the molarity of a solution, we can calculate the amount of solute contained in a specified volume of that solution. Calculate the volume in litters of a M solution of sulphuric acid, H 2 SO 4, required to react completely with grams of Na 2 CO 3 according to the equation The balanced equations tells us For the reaction to go to completion

31 Find the volume in litres and in millilitres of a M NaOH solution required to react with 40.0 mL of M H 2 SO 4 solution according to the reaction The balanced equation tells us that the reaction ratio is 1 mol of H 2 SO 4 to 2 mol of NaOH


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