Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 10 Mass relationships in chemical reactions: Stoichiometry.

Similar presentations


Presentation on theme: "Chapter 10 Mass relationships in chemical reactions: Stoichiometry."— Presentation transcript:

1 Chapter 10 Mass relationships in chemical reactions: Stoichiometry

2 To Review How many grams in a mole of CaSO 4 Al 2 (SO 4 ) 3 Ca 1 x 40.1 g/mol = 40.1 g S 1 x 32.1= 32.1 O 4 x 16.0 = g/mol Al 2 x 27.0 g/mol = 24.0 S 3 x 32.1 = 96.3 O 12 x 16.0 = g/mol

3 Al(s) + 3Br 2 (l) 1Al 2 Br 6 (s) 2 Al(s) + 3Br 2 (l) 1Al 2 Br 6 (s) Stoichiometric Coefficients- the coefficients (number in front of the chemical formula) in a balanced chemical equation Stoichiometry- the relationship between the quantities of chemical reactants and products Stoichiometric factor- a mole ratio relating moles of one substance to another substance involved in the same chemical reaction

4 Create two stoichiometric factors that relate the amount of the reactant H 2 O 2 to the product O 2 2 H 2 O 2 (liq) ---> 2 H 2 O(l) + 1 O 2 (g) 2 H 2 O 2 (liq) ---> 2 H 2 O(l) + 1 O 2 (g) 2 mol H 2 O 2 OR 1 mol O 2 1 mol O 2 2 molH 2 O 2 1 mol O 2 2 molH 2 O 2

5 Create two stoichiometric factors that relate the amount of the product H 2 O to the reactant H 2 O 2 2 H 2 O 2 (l) ---> 2 H 2 O(l) + O 2 (g) 2 H 2 O 2 (l) ---> 2 H 2 O(l) + O 2 (g) 2 mol H 2 O 2 OR 2 mol H 2 O 2 mol H 2 O 2 OR 2 mol H 2 O 2 mol H 2 O 2 mol H 2 O 2 2 mol H 2 O 2 mol H 2 O 2

6 Create two stoichiometric factors that relate the amount of the product CO 2 to the reactant C 4 H 10 2 C 4 H O > 10H 2 O + 8CO 2 8 mol CO 2 OR 2 mol 8 mol CO 2 OR 2 mol C 4 H 10 2 mol 8 mol CO 2 2 mol C 4 H 10 8 mol CO 2

7 Stoichiometric Calculations The most common types of equation stoichiometry problems are called 3 step problems  Given the mass of a reactant or product determine the equivalent mass of a different reactant or product  These are called mass to mass or gram to gram conversions  Consider: 2 H 2 O > 2 H 2 + O 2 How many g of H 2 are produced by decomposing 1.00 g H 2 O?

8 Stoichiometric Calculations Mass H 2 O Mol H 2 O Mol H 2 Mass H 2 1 mol MM in g 1 mol mol H 2 mol H 2 O All calculations are done using conversion factors Each arrow represents a conversion factor Align units first, then place numbers into the conversion factor

9 Stoichiometric Calculations Step 1 - Convert g of given reactant or product to moles  Use molar mass of the given Step 2 - Convert from moles of the given to moles of the unknown  Use stoich factor as conversion factor Step 3 - Convert from moles to g of the unknown  Use molar mass of the unknown as conversion factor

10 Problem 1a. If 37.6 g of water is decomposed to hydrogen and oxygen, how many grams of hydrogen will be produced? Write the balanced chemical equation 2 H 2 O > 2 H 2 + O 2

11 Write the information that is given and what's asked for above the equation Mass37.6 g ? g Equation2H 2 O > 2H 2 + O 2

12 Write the molar mass of the substances. Mass37.6 g ? g Equation2 H 2 O > 2H 2 + O 2 Molar Masses (g/mol) Remember the Pathway: g H 2 O mol H 2 O mol H 2 g H 2  

13 Step 1 - Convert the given mass of water to moles of water 1 mol H 2 O = 18.0 g H 2 O 37.6 g H 2 O x 1 mol H 2 O 18.0 g H 2 O 2 H 2 O > 2 H 2 + O 2

14 Step 2 - Convert moles of water to moles of hydrogen using stoichiometric factor 2 H 2 O > 2 H 2 + O g H 2 O x 1 mol H 2 O x 2 mol H g H 2 O 2 mol H 2 O

15 Step 3 - Convert moles of hydrogen to mass of hydrogen 1 mol H 2 = 2.02 g H g H 2 O x 1 mol H 2 O x 2 mol H 2 x 2.0g H 2 = 4.17g H g H 2 O 2 mol H 2 O 1 mol H 2 2 H 2 O > 2 H 2 + O 2

16 Problem 1b. At room temperature and pressure, aluminum reacts with oxygen to give aluminum oxide. If you react 161 g of Al, what mass of O 2 is needed for complete reaction? Write the chemical equation: Al(s) + O 2 (g) -----> Al 2 O 3

17 2 You need a balanced chemical equation: Al(s) + O 2 (g) -----> Al 2 O

18 Mass161 g ? g Equation4Al(s) + 3O 2 (g) ---->2Al 2 O 3 Molar Masses (g/mole) Record given mass and what is asked for in the problem

19 3 Step Calculations 161 g Al x 1 mol Al 27.0 g Al x 3 mol O 2 4 mol Al x 32.0 g O 2 1 mol O 2 4Al(s) + 3O 2 (g) ---->2Al 2 O 3 = 143 g O g Al ? g O g/mol 32.0 g/mol g Al  mol Al  mol O 2  g O 2

20 What mass of MgO can be produced from igniting 1.5 g of Mg in oxygen? Mass 1.5 g ? g Equation 2 Mg(s) + O 2 (g) ---> 2 MgO Molar Masses (g/mole) 1.5 g Mg x 1 mol Mg x 24.3 g Mg 2 mol MgO x 2 mol Mg 40.3 g MgO 1 mol MgO = 2.49 g MgO

21 Theoretical vs Actual Yield Theoretical yield = the amount of product that could be produced if the reaction is perfect  The amount that results from the 3 step process  This never happens  You usually get less than the theoretical yield Actual yield = The amount of product you recover when you run the experiment

22 Limiting and Excess Reactant In chem reactions, reactants are never present in the exact ratio to ensure they are completely used up in the reaction You usually run out of one reactant, and the other is in excess (left over)  Limiting reactant = the one completely used up  Excess reactant = the one with some left over

23 A Chemical Reaction Reactants Products All of is used up = Limiting Reactant Some of is left over = Excess Reactant

24 Limiting Reactant Problem Determine the limiting and excess reactant if 1.40 g of N 2 reacts with 1.00 g of H 2 to produce ammonia, NH 3. N 2 + 3H 2  2NH g 1.00 g 28.O g/mol 2.0 g/mol 1.40 g N 2 x 3 mol H 2 1 mol N 2 x 2.0 g H 2 1 mol H 2 x 1 mol N g N 2 = 0.30 g H 2 There is more than 0.30 g of H 2 present, therefore H 2 will be left over (is in excess) and N 2 will be the limiting reactant

25 Percent Yield A means of quantifying how efficient a chemical reaction is % Yield = Actual Yield Theoretical Yield x 100 % Yields can be above 100% What’s better, a % yield of 90% or 110% ? % Yields are usually below 100% less product is recover than is possible

26 Solving Percent Yield Probs  1 – Determine the limiting reactant 2 – Calculate the theoretical yield from the given amount of the limiting reactant 3 – Calculate the % yield (the actual yield needs to be given to you in the problem). % Yield = Actual Yield Theoretical Yield x 100

27 Percent Yield Practice Prob # g N 2 x 3 mol H 2 2 mol N 2 x 2.0 g H 2 1 mol H 2 x 1 mol N g N 2 = g H 2 A student reacts 6.00 g of N 2 reacts with g of H 2 to produce ammonia, NH 3. if 2.59 g of NH 3 is recovered from the reaction, what is the student’s % yield? N 2 + 3H 2  2NH 3 Therefore, g H 2 is the limiting reactant Step 1 Determine the limiting reactant.

28 Percent Yield Practice Prob # g H 2 x 2 mol NH 3 3 mol H 2 x 17.0 g NH 3 1 mol NH 3 x 1 mol H g H 2 = 2.83 g NH 3 Step 2 Deter the theoretical yield from limiting reactant Actual yield = 2.59 g NH 3 (given in the problem) Theoretical yield = 2.83 g NH 3 % yield = actual yield x 100 theroretical yield = 2.59 g x 100 = 91.5% 2.83 g

29 Lab Exercise A student reacts 2.40 g of NaHCO 3 with an excess of HCL according to the reaction: 2.40 g X g NaHCO 3 + HCl ---> NaCl + H 2 O + CO g/mol 58.5 g/mol Determine the theoretical yield of NaCl (3 step process) g NaHCO 3  mol NaHCO 3  mol NaCl  g NaCl x 1 mol NaHCO g NaHCO 3 x 1 mol NaCl 1 mol NaHCO 3 x 58.5 g NaCl = 1 mol NaCl 2.40 g NaHCO 3 = 1.67 g NaCl

30 Lab Exercise The reaction from your lab is conducted with a 2.40 g sample of NaHCO 3. In the lab, the reaction produces 1.57 g of NaCl. What is the percent yield? x 1 mol NaHCO g/mol x 1 mol NaCl 1 mol NaHCO 3 x 58.5 g NaCl = 1 mol NaCl 2.40 g NaHCO g NaCl % Yield = actual yield x 100 theoretical yield = 1.57 g x g = 94.0%


Download ppt "Chapter 10 Mass relationships in chemical reactions: Stoichiometry."

Similar presentations


Ads by Google