Mole Ratio - used to predict what would happen if the amount of one of these substances is changed. Eg. In above - How many moles of O2 would react with 1 mole of Mg to form MgO? Ratio = 2 mol Mg = 1 mol Mg 1 mol O2 x mol O2 x mol O2 = 1 mol O2 x 1 mol Mg 2 mol Mg = 0.5 - How many moles of MgO could be produced from 1 mole Mg? Ratio = 2 mol MgO = x mol MgO = 1 mol 2 mol Mg 1 mol Mg
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Instant Practice – How many moles of MgO would be produced from 0.175 moles O2? Given: 2 Mg + 1 O2 Æ 2 MgO 1. Mole Ratio 2:1:2 2 mol MgO = x mol MgO 1 mol O2 0.175 mol O2
= 0.35 mol
Question: How many g of MgO? 0.35 mol x
40.3 g MgO mol
= 14.11g
6-3 Grams Æ Moles Æ Atoms How many molecules of MgO would be produced from the oxidation of 2.00g of Mg? Given: 2.00g Mg
5) Convert mol MgO to molecules 0.0823 mol x 6.02 x 1023 molecules U = 4.95 x 1022 mol Exercises p.154 #2,3,4
6-4 Stoichio - metry - elements - measurements - calculating or measuring elements or compounds involved in a chemical change Exercises p. 154 Review #1,2 on board; #3-6 instant practice 6-5 Mole - Mass calculations Example 6.2 How many moles of sodium metal would be needed to react with chlorine gas to make 735g of sodium chloride? 1) Write and balance equation 2 Na(s) + Cl2(g) Æ 2 NaCl(s) 2) mol ratio = 2:1:2 3) Find mol NaCl 737g x 1 mol = 12.60 mol 58.5g NaCl 4) Using mol ratio determine mol Na 3
2 mol Na = x mol Na = 12.60 mol Na 2 mo NaCl 12.60mol NaCl 5) Determine mass Na 12.60 mol Na x 23.0 g Na = 289.8g mol sig figs - 290g 6-6 Mass - Mass Calculation How many grams of potassium chloride, KCl, are produced by decomposing 118g of potassium chlorate, KClO3? 1) Write balanced equation KClO3 Æ KCl + 3O 2) Mol ratio 2: 1: 3 3) Determine mol of KClO3 118g KClO3 x 1 mol = 0.96 mol 122.6g KClO3 Side calc. H - 1 x 39.1 = 39.1 Cl - 1 x 35.5 = 35.5 O - 3 x 16.0 = 48.0 122.6 2 KClO3 Æ 2 KCl + 3 O2 0.96 mol Æ ?? 4) Using mol ratio - determine mol KCl 2 mol KCl = x mol KCl = 0.96 mol KClO3 2 mol KClO3 0.96 mol KClO3 5) Determine g KCl
4
0.96 mol KCl x
6-7
74.6 g mol
= 71.6 g KCl
Molarity and Replacement
Review: molarity (M) = moles solute 1 Litre solution so… Given: 250mL of 3.0M AgNO3(aq) (or any other) = 3.0 moles = x moles = 1 Litre 0.250 L 3.0 moles x 0.250L = 0.75 moles ???Okay???
Problem: How many grams of copper will react to completely replace sliver from 0.100M solution of silver nitrate, AgNO3? 1) Write balanced equation Cu + AgNO3 Æ Cu(NO3)2 + 2 Ag 2) Mol ratio 1:2:1:2 3) Determine mol AgNO3 0.100 mol AgNO3 = x mol AgNO3 = 0.021 mol 1L 0.208 L
4) Determine mol Cu (using mol ratio) 5
1 mol Cu = x mol Cu 2 mol AgNO3 0.0208 mol AgNO3
= 0.0104
5) Determine g Cu 0.0104 Mol Cu x
6-8
63.5 g mol
= 0.660g
Limiting reagents (reactants) and reactants “in Xs” - an excess reactants is one that is added in larger then necessary amounts
- limiting reactants determines (or limits) the amount of products Eg. Zn + Given: 1.21mol
2HCl
Æ
ZnCl2
+
H2
2.65mol
Find: Which reactant is in excess? How much in excess? How many g ZnCl2 produced? How many g H2 produced? 1) What is the mole ratio? 1 Zn : 2 HCl 2) Determine how much HCl would be required to react with 1.21 mol Zn (using mol ratio)?
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2 mol HCl = x mol HCl = 2.42 mol HCl 1 mol Zn 1.21 mol Zn 3) Determine excess HCl 2.65 HCl - 2.42 mol HCl = 0.23 mol NXS 4) If HCl is NXS the Zn must be limiting reactant 5) Determine moles ZnCl2 using mol ratio 1 mol ZnCl2 = x mol Zn Cl2 = 1.21 mol ZnCl2 1 mol Zn 1.21 mol Zn 6) Determine moles H2 using mol ratio 1 mol H2 = x mol H2 = 1.21 mol H2 1 mol Zn 1.21 mol Zn
Example 6.9 When 79.1 g of Zinc are reacted with 1.05L of 2.00M hydrochloric acid, HCl, to produce zinc chloride, ZnCl2, and hydrogen gas, H2, which reactant will be in excess and by how much? Calculate mass of each product.
4. Determine moles of HCl available 2.00 moles = x moles = 2.10 mol 1L 1.05L 5) Determine mol ratio Zn + 2 HCl Æ ZnCl2 + H2 1 2 6. a) Determine moles of HCl needed to react with 1.21 mol Zn. (using mol ratio) 2 mol HCl = x mol HCl = 2.42 mol HCl 1 mol Zn 1.21 Zn * 2.42 mol HCl is needed to react with 1.21 mol Zn. However, there is only 2.10 mol HCl available. HCl is the limiting reactant.
6. b) Determine moles of Zn needed to react with 2.10 mol HCl. 1 mol Zn = x mol Zn = 1.05 mol 2 mol HCl 2.10 mol HCl * There is 1.21 mol Zn available but we only need 1.05 mol Zn to react with 2.10 mol HCl. Therefore, 1.21g - 1.05g Zn = 0.16 mol zinc is in XS.
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7. Rewrite balanced equation noting mole ratio and mol of reactants involved in actual Rxn. Zn + 2HCl Æ ZnCl2 + H2 1 : 2 : 1 :1 8. a) Determine mol ZnCl2 formed using mol ratio b) Determine g ZnCl2 1.05 mol x
136.4 g mol
= 143.2g
9. a) Determine mol H2 formed using mol ratio 1.05 mol x 2.02g H2 = 2.12g mol
Percent Yield - Rarely…does a reaction work in the lab where you get the amount of product that you think you should. Usually, it is far less. This is expressed in Percent Yield. - Theoretical Yield - what the balanced equation says you should get - Actual Yield - what you really get - Percent Yield -
__ actual __ theoretical
x100%
Example 6-10
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When 45.8g of K2CO3 are reacted completely with excess HCl, 46.3g KCl are produced. H2O and CO2 is also formed. Calculate theoretical and percent yield of KCl. 1. Write balanced equation K2CO3 + HCl Æ KCl + H2O + CO2 mol ratio 1
: 2
: 2
:
1
: 1
2. Determine mol K2CO3 45.8g K2CO3 x 1 mol = 0.33 mol 138.2g K2CO3 K C O
Side calc 2 x 39.1 = 78.2 1 x 12.0 = 12.0 3 x 16.0 = 48.0 138.2
4. Determine Percent Yield Actual (given) x 100% = 46.3g x 100% = 94% Theoretical (calc) 49.24g
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How to Page Solving Dilution Problems There are two ways to go about this: C1v1=c2v2 Or Rationalize your way through. The following is dedicated to the latter method: Steps 1) Find moles of solute required in final solution 2) Apply molar ratio: moles (start) = x moles (finish) L (start) xL (finish) 3) Solve for X Eg. Describe how you would prepare a 750mL quantity of 0.10M NaIO3 from a 0.20M 1. Moles NaIO3 in final volume 0.10 moles x 0.750L = 0.075 moles 1L (find M) (find V) (find mol of solute req’d) 2. 0.20 moles = 0.075 moles L x Litres (flip) L = x Litres = 1 L x 0.075 mol = 0.375 L 0.02 moles 0.075 moles 0.20 mol or: 375 mL To 375mL of stock solution, add enough water to make 750mL ( ie
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750mL - 375mL = 375mL Vf
- stock solution = water
The How to Sheet Solving Mole Ratio Problems 1. Moles to Grams Problem: How many grams of MgO would be produced by the oxidation of 0.175 moles of O2 magnesium? Given: 0.175 moles of O2 Æ grams MgO 1) Write and balance equation 2Mg + O2 Æ 2MgO 2) Determine mole ratio 3) Plot a road map
2Mg
+
O2 0.175 moles O2 (convert)
a)
Æ Æ
2MgO grams MgO convert)
moles O2Æ mol ratio Æ moles MgO
a) mole ratio 2 mol MgO = x mol MgO 1 mol O2 0.175 mol O2 x = 0.35 mol MgO
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b) convert 0.35 mol x 40.3 g Mg = 14.11g mol
2. Grams to grams Problem: How many grams of MgO would be produced from the oxidation of 2.00g of Mg? Given: 2.00g Mg Æ x g MgO 1) Write and balance equation
2Mg
+
O2
Æ
2MgO
O2
Æ
2MgO
Æ
g MgO
2) Determine mol ratio 3) Plot a road map
2Mg
+
2.00g (convert)
Æ
moles Mg
moles MgO
a) Convert 2.00g Æ mol Mg 2.00g x
1 mol = 0.082 mol Mg 24.3g Mg
b) Mole ratio Mg Æ MgO 2 mol MgO = x mol MgO
x = 0.082 mol 13
2 mol Mg
0.082 mol Mg
c) Convert mol MgO Æ g MgO 0.082 mol MgO x 40.0g MgO = 3.29g MgO mol 3. Grams to molecules Problem: How many molecules of MgO would be produced from the oxidation of 2.00g of Mg? Given: 2.00g Mg Æ molecules MgO
1) Write and balance equation
2) Determine the mol ratio 3) Plot a road map 2.00g Mg
Æ
molecules MgO
a) convert
c) convert
mol Mg Æ b) mol ratio Æ
mol MgO
a) Convert 2.00g Æ mol Mg 2.00g x 1 mol = 0.082 mol Mg 24.3g Mg b) Mol ratio Mg Æ MgO 2 mol MgO = x mol MgO 2 mol Mg 0.082 mol Mg
x = 0.082 mol
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c) Convert mol MgO Æ molecules MgO 0.082 mol MgO x 6.02 x 1023 molecules mol = 4.94 x 1022 molecules Mol - Mass* *= pg. 164 (1-8) Use Road Map Mass - Mass*
More How-To For Molarity and Replacement Reactions Example 6.6 The good old Silver Nitrate replacement reaction! Given: 208mL of 0.100M AgNo3(aq) Find: g Cu to completely replace AgNo3(aq) 1) Write and balance equation
2AgNo3(aq)
+
Cu(s) -->Cu(NO3)2 (aq) + 2Ag(s)
+
Cu(s) -->Cu(NO3)2 (aq) + 2Ag(s)
2) Road Map
2AgNo3(aq)
208mL 0.100M AgNo3(aq) a)
Æ g Cu c)
15
mol AgNo3(aq)
Æ b)
mol ratio Æ mol Cu
a) 208mL 0.100 M Æ mol AgNo3(aq) 0.208L x 0.100M = 0.0208 mol L b) 2 mol AgNo3(aq) Æ 1 mol Cu 2 mol AgNo3(aq) = 0.0208 mol AgNo3(aq) 1 mol Cu x mol Cu Æ Flip and multiply 1 mol Cu x 0.0208 mol AgNo3(aq) = x mol Cu = 0.0104 mol 2 mol AgNo3(aq) c) mol Cu Æ g Cu
0.0104 mol x 63.5g = 0.660g mol
Molarity Flavor #1 - Given grams of a solute in a volume of solution - find molarity Problem: What is the molarity of a 125mL solution containing 25g of silver nitrate ( AgNo3 )
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(c)
(b)
Molarity = moles Litre
Å 25g AgNo3 (a) molecular wt AgNo3 from PT
Given: 125mL = 0.125L a) Molecular Wt. AgNo3
b) Moles AgNo3
Ag N O
= 25g AgNo3 169.9 g/mol AgNo3
1x107.89 1x14.01 3x16.00 169.9
= 0.147 mol c) Molarity = 0.147 mol 0.125 L
= 1.18 M
Flavor #2 - Given the molarity of a solution and a volume - find “corresponding” grams
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Problem: How many grams of copper sulphate ( CuSO4 ) are in 75mL of a 5 M solution? 1) Set up a molar ratio (5 M CuSO4) 5 mol CuSO4 = IL
x mol CuSO4 0.075L
x = 0.376 moles CuSO4 (in 75mL of CuSO4) 2) Convert moles to grams molecular Wt. CuSO4 i) Cu 1x63.55 = 63.55 S 1x32.06 = 32.06 O 4x16.00 = 64.00 160.61
ii) grams = mol x molecular Wt. = 0.375 x 160.61g/mol = 60.23 g or 60 g (Sig Figs)
Flavor #3 - Given the molarity of a solution and the mass of a solute find corresponding volume
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Problem: What volume of a 1.5M solution would you find 3 grams of sodium chloride? 1) Set up molar ratio 1.5M NaCl 1.5 mol NaCl = moles NaCl 1L xL 2) Convert grams to mol
Å(3g)
i) Na 1x22.99 = 22.99 Cl 1x35.45 = 35.45 58. 44
ii) moles =
grams NaCl molecular wt. NaCl = 0.05 mol
iii) 1.5 mol NaCl = 0.05 mol NaCl 1L xL * flip and solve for X = 0.075 L (75mL)
