Chemical Equations
Chemical Equations shows the results of a chemical process reactants (reagents) products coefficients the numbers in front of formulas in chemical equations gives the relative number of molecules taking part in a reaction
2H (g) 2 + O (g) 2 2H 2 O(g) Chemical bonds have been broken and new chemical bonds have been formed
Writing Chemical Equations 2H 2 (g) + O 2 (g) 2H 2 O (g) + 2 moles 1 mole 2 moles 4.04 g 32.00 g 36.04 g Parentheses show physical state of substances
Eample molar interpretation H 2 (g) + Cl 2 (g) 2HCl(g) Start: 1 mol 1 mol 0 Finish: 0 0 2 mol
Eample mass interpretation 2H 2 + O 2 2H 2 O 4 g 32 g 36 g Start: 4g 32g 0 Finish: 0 0 36 g
The law of conservation of mass requires that chemical equations must balance. What goes in Must come out
Quantitative Relationships
Stoichiometry the study of quantitative between amounts of reactants used and amounts of products formed how much reactant is needed to yield a certain amount of product?
Types of problems Mole - mole calculations Mass - mass calculations Limiting/Ecess reagent calculations
The mole method 1. Write and balance the equation. 2. Convert the given quantities into moles. 3. Use the coefficients in the balanced equation to relate the number of moles of known substances to the desired unknown one. 4. Convert to desired units. 5. Check your answer.
The mole method 1. Write and balance the equation. 2. Convert the given quantities into moles. 3. Use the coefficients in the balanced equation to relate the number of moles of known substances to the desired unknown one. 4. Convert to desired units. 5. Check your answer.
Stoichiometry Molar ratio of Y to X n y n Moles of X Moles of Y Mass of X X Y Mass of Y
Eample How many grams of nitrogen dioide can be formed by reaction of 1.44 g of nitrogen monoide with oygen? 2NO + O 2 2NO 2
Stoichiometry Molar ratio of Y to X n y n Moles of X Moles of Y 1.44 g of NO Mass of NO 2 2NO + O 2 2NO 2
Stoichiometry Molar ratio of Y to X n y n 0.048 Moles NO Moles of Y 1.44 g of NO Mass of NO 2 2NO + O 2 2NO 2
Stoichiometry Molar ratio NO 2 to NO 2 2 0.048 Moles NO Moles of Y 1.44 g of NO Mass of NO 2 2NO + O 2 2NO 2
Stoichiometry Molar ratio NO 2 to NO 2 2 0.048 Moles NO 0.048 Moles NO 2 1.44 g of NO Mass of NO 2 2NO + O 2 2NO 2
Stoichiometry Molar ratio NO 2 to NO 2 2 0.048 Moles NO 0.048 Moles NO 2 1.44 g of NO 2.21 g of NO 2 2NO + O 2 2NO 2
Eample 2NO + O 2 2NO 2 1.44g NO 1mol NO 30g NO 2mol NO 2 2mol NO 46g NO 2 1mol NO 2 = 2.21g NO 2
pg. 359 Problem 10a Calculate the number of moles of CS2 produced when 1.50 mol S 8 is used. 2 CH4 + S 8 2CS 2 + 4 H 2 S 1.5 mol S 8 2 mol CS 2 = 3 mol CS 2 1 mol S 8
pg. 359 Problem 10b Calculate the number of moles of H2S produced when 1.50 mol S 8 is used. 2 CH4 + S 8 2CS 2 + 4 H 2 S 1.5 mol S 8 4 mol H 2 S = 6 mol H 2 S 1 mol S 8
pg. 360 Problem 11 Titanium tetrachloride (TiCl4) is etracted from titanium oide using chlorine and carbon. TiO 2 + C + 2 Cl 2 TiCl 4 + CO 2 if you begin with 1.25 mol TiO2, what is the mass of Cl 2 needed? 1.25 mol TiO 2 2 mol Cl 2 70.9 g Cl 2 1 mol TiO 2 1 mol Cl 2 = 177 g Cl 2
pg. 362 Problem 13 Air bags in cars can be inflated using the decomposition of sodium azide (NaN 3 ). 2 NaN 3 2 Na + 3 N 2 determine the mass of N2 produced if 100.0 g of NaN 3 is decomposed? 100.0 g NaN 3 1 mol NaN 3 3 mol N 2 65.0 g NaN 3 2 mol NaN 3 28 g N 2 1 mol N 2 = 64.64 g N 2
Limiting Reagents
Limiting Reagent Reactants are not always present (or available) in stoichiometric quantities. One reactant may be present in quantities such that it is completely consumed while ecess amounts of other reactants remain. - called limiting reactant or limiting reagent The limiting reagent will limit the amount of product produced.
Eample How many moles of MgCl 2 will be produced? Mg + Cl 2 MgCl 2 Start 1 mol 1 mol 0 Finish 0 0 1 mol
Eample How many moles of MgCl 2 will be produced? Mg + Cl 2 MgCl 2 Start 1 mol 2 mol 0 Finish 0 1 mol 1 mol magnesium is the limiting reagent 1 mol of chlorine will be left unchanged
Limiting Reagent Molar ratio Y to X n y n Moles of X Moles of W Moles of Y Mass of X Mass of W W + X Y Mass of Y
Compare molar ratio W to X to their coefficients in balanced equation; identify LR Molar ratio Y to LR Moles of X Moles of W Moles of Y Mass of X Mass of W W + X Y Mass of Y
Eample Determine the limiting reagent and the amount of PI 3 produced when 6.00g P 4 reacts with 25.0g of I 2. P 4 + 6I 2 4PI 3 1mol P 6.00g P 4 4 =.0484mol P 4 124g P 4 1mol I 25.0g I 2 2 =.0984mol I 2 254g I 2
Eample cont... Determine how much I 2 would be needed to react completely with the available amount of P 4. if I have but I only have P 4 + 6I 2 4PI 3.0484mol P 4 6mol I 2 1mol P 4 = I need 0.290mol I 2.0984mol I 2 I 2 is the limiting reagent
Eample cont... the amount of PI 3 produced from the limiting reagent... P 4 + 6I 2 4PI 3.0984mol I 2 4mol PI 3 6mol I 2 412g PI 3 1mol PI 3 = 27.0g PI 3
pg. 368 problem 20a the reaction between solid sodium and iron (III) oide is one in a series of reactions that inflates an automobile air bag Fe 2 O 3? 6Na + Fe 2 O 3 3Na 2 O + 2Fe if 100.0 g Na and 100.0 g Fe 2 O 3 are used in this reaction, determine: 100.0 g Na a. the limiting reactant. 1 mol Na 23.0 g Na = 4.348 mol Na 1 mol Fe 2 O 3 100.0 g Fe 2 O 3 159.9 g Fe 2 O 3 = 0.6254 mol Fe 2 O 3
pg. 368 problem 20a if 100.0 g Na and 100.0 g Fe 2 O 3 are used in this reaction, determine: a. the limiting reactant. 6Na + Fe 2 O 3 3Na 2 O + 2Fe if I have 1 mol Fe 4.348 mol Na 2 O 3 = 0.724 mol Fe2 O 3 6 mol Na I need 0.6254 mol Fe 2 O 3 is limiting but I only have
pg. 368 problem 20a if 100.0 g Na and 100.0 g Fe 2 O 3 are used in this reaction, determine: b. the reactant in ecess. 6Na + Fe 2 O 3 3Na 2 O + 2Fe 4.348 mol Na in ecess 1 mol Fe 2 O 3 = 0.724 mol Fe2 O 3 6 mol Na needed 0.6254 mol Fe 2 O 3 available
pg. 368 problem 20a if 100.0 g Na and 100.0 g Fe 2 O 3 are used in this reaction, determine: c. the mass of iron produced. 6Na + Fe 2 O 3 3Na 2 O + 2Fe 2 mol Fe 55.85 g Fe 0.6254 mol Fe 2 O 3 1 mol Fe 2 O 3 1 mol Fe = 69.86 g Fe produced
pg. 368 problem 20a if 100.0 g Na and 100.0 g Fe 2 O 3 are used in this reaction, determine: 6Na + Fe 2 O 3 6 mol Na 0.6254 mol Fe 2 O 3 1 mol Fe 2 O 3 = 86.30 g Na used d. the mass of ecess reactant. 3Na 2 O + 2Fe 23.0 g Na 1 mol Na 100.0 g Na - 86.30 g Na = 13.7 g Na left over
work sheet eample 3CaCO 3 + 2FePO 4 Ca 3 (PO 4 ) 2 + Fe 2 (CO 3 ) 3 if 100g calcium carbonate and 45 g iron(iii) phosphate are used in this reaction, determine: a. the limiting reactant. 100 g CaCO 3 1 mol CaCO 3 100 g = 1 mol CaCO3 1 mol FePO 4 45 g FePO 4 151 g = 0.298 mol FePO 4
work sheet eample 3CaCO 3 + 2FePO 4 is limiting Ca 3 (PO 4 ) 2 + Fe 2 (CO 3 ) 3 if 100g calcium carbonate and 45 g iron(iii) phosphate are used in this reaction, determine: a. the limiting reactant. 1 mol CaCO 3 if I have 0.298 mol FePO 4 but I only have 2 mol FePO 4 = 0.66 mol Fe2 O 3 3 mol CaCO 3 I need
work sheet eample 3CaCO 3 + 2FePO 4 Ca 3 (PO 4 ) 2 + Fe 2 (CO 3 ) 3 if 100g calcium carbonate and 45 g iron(iii) phosphate are used in this reaction, determine: b. grams of Ca 3 (PO 4 ) 2 formed? 1 mol Ca 0.298 mol FePO 3 (PO 4 ) 2 4 2 mol FePO 4 310 g 1 mol Ca 3 (PO 4 ) 2 = 46.2 g Ca 3 (PO 4 ) 2 produced
work sheet eample 3CaCO 3 + 2FePO 4 Ca 3 (PO 4 ) 2 + Fe 2 (CO 3 ) 3 if 100g calcium carbonate and 45 g iron(iii) phosphate are used in this reaction, determine: b. grams of Fe 2 (CO 3 ) 3 formed? 1 mol Fe 0.298 mol FePO 2 (CO 3 ) 3 4 2 mol FePO 4 291.7 g 1 mol Fe 2 (CO 3 ) 3 = 43.5 g Fe 2 (CO 3 ) 3 produced
work sheet eample 3CaCO 3 + 2FePO 4 Ca 3 (PO 4 ) 2 + Fe 2 (CO 3 ) 3 if 100g calcium carbonate and 45 g iron(iii) phosphate are used in this reaction, determine: c. grams of CaCO 3 unreacted? 3 mol CaCO 0.298 mol FePO 3 4 2 mol FePO 4 100 g 1 mol CaCO 3 = 44.7 g CaCO 3 used 100 g - 44.7 g = 55.3 g CaCO 3 unused
Eample From the reaction between of 10.0g of Hg and 9.0g of Br 2. What mass of which reagent is left unreacted? Hg + Br 2 10.0g Hg 1 molhg 200.6gHg = HgBr 2 Hg is limiting 4.99 10-2 molhg 9.0g Br 2 1 molbr 2 = 5.63 10-2 molbr 2 159.8gBr 2
Hg + Br 2 HgBr 2 4.99 10-2 molhg 1 molbr 2 = 4.99 10-2 molbr 2 1 molhg moles of Br 2 needed to use up Hg available 159.8gBr 4.99 10-2 molbr 2 2 1 molbr 2 = 7.97 g Br 2 grams of Br 2 used 9.0g Br 2-7.97 g Br 2 = 1.03 g Br 2 ecess
Reaction Yield
Theoretical yield the amount of product that would result if all the limiting reagent reacted Actual yield the amount of product actually obtained from the reaction Almost always less than the theoretical yield
Percent Yield Actual yield %yield = 100% Theoretical yield Determines how efficient a reaction is
Eample In a certain industrial operation 3.54 10 7 g of TiCl 4 is reacted with 1.13 10 7 g of Mg. (a) Calculate the theoretical yield of Ti in grams. (b) Calculate the percent yield if 7.91 10 6 g are actually obtained. TiCl 4 + 2Mg Ti + 2MgCl 2
Calculate theoretical yield 3.54 10 7 g TiCl 4 25.0g Mg 1mol TiCl 4 = 1.87 10 5 mol 187.7g TiCl 4 TiCl 4 I have 1mol Mg = 4.65 10 5 mol 24.31g Mg Mg if I have 2mol Mg 1.87 10 5 mol TiCl 4 = 3.74 10 5 mol Mg 1mol TiCl 4 I need there is more than enough Mg TiCl 4 is limiting
3.54 10 7 g TiCl 4 1mol TiCl 4 1mol Ti 187.7g TiCl 4 1mol TiCl 4 47.88g Ti 1mol Ti = 7.91 10 6 g Ti 100% = 8.93 10 6 g Ti 8.93 10 6 g Ti Actual yield %yield = 100% = 88.6% theoretical Theoretical yield
another method Determine the limiting reagent and the amount of PI 3 produced when 6.00g P 4 reacts with 25.0g of I 2. P 4 + 6I 2 4PI 3 6.00g P 4 1mol P 4 124g P 4 =.0484mol P 4 25.0g I 2 1mol I 2 254g I 2 =.0984mol I 2
another method P 4 + 6I 2 4PI 3.0984mol I 2 = 2.033.0484mol P 4 The actual I 2 /P 4 ratio is less than the stiochiometic ratio 6 mol I 2 = 6 So there is not enough I2 1 mol P to 4 I 2 is the limiting reagent react with all the P 4