Lecture Notes Week 01

Lecture Notes Week 01 - CHEM*1040 WEEK 1 Review Questions:...

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1 CHEM*1040 – WEEK 1 Review Questions : 1) How many H atoms are there in 45.0 g (NH 4 ) 2 Cr 2 O 7 ? Given: MM N = 14.01; MM H = 1.008; MM Cr = 52.00; MM O = 16.00 A) 2.15 × 10 23 atoms B) 6.02 × 10 23 atoms C) 7.52 × 10 23 atoms D) 8.60 × 10 23 atoms 2) How many moles of calcium are required to produce 5.50 g of calcium phosphate? The molar mass of calcium phosphate is 310.18 g/mol A) 1.77 × 10 –2 mol B) 5.32 × 10 –2 mol C) 6.28 × 10 –2 mol D) 4.07 × 10 –2 mol 3) Combustion analysis of a 12.0–mg sample of ibuprofen produced 9.50 mg of water. What is the percent hydrogen in ibuprofen? A) 8.87% B) 17.7% C) 4.44% D) 79.2% Answers to the above questions: 1 d); 2 b); 3 a) CHEMISTRY’S ACCOUNTING Reaction stoichiometry allows us to use balanced equations to convert/determine the following: Mass Moles Molecules Atoms What do balanced equations tell you? e.g., 3H 2 + N 2 2NH 3 molecules? moles? masses? atoms?
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2 Example 1 : What mass of I 2 is produced if 13.1 g KI is reacted with excess KIO 3 and HNO 3 according to the following balanced equation? (Ans: 12.0 g) Given: MM(KI) = 166.0 g mol 1 and MM(I 2 ) = 253.8 g mol 1 Rxn: 5KI(aq) + KIO 3 (aq) + 6HNO 3 (aq) 6KNO 3 (aq) + 3I 2 (aq) + 3H 2 O(l) Strategy : 1) Convert mass of KI to moles of KI – using the molar mass of KI. 2) Convert moles of KI to moles of I 2 – using the stoichiometry of the balanced equation. 3) Convert moles of I 2 to mass of I 2 – using the molar mass of I 2 . Example 2: PCl 3 is prepared by a two step reaction: 4HCl(aq) + MnO 2 (s) MnCl 2 (aq) + 2H 2 O(l) + Cl 2 (g) P 4 (s) + 6Cl 2 (g) 4PCl 3 (s) How many moles of PCl 3 will be prepared from 0.2743 moles HCl and excess MnO 2 and P 4 ? (Ans: 0.04572 moles) Strategy : Convert moles of HCl to moles of PCl 3 using the stoichiometry of the balanced equations: a. moles of HCl to moles of Cl 2 (an intermediate which connects the two reactions) b. moles of Cl 2 to PCl 3 In general, unless we are TOLD that the other reagents are in EXCESS , we must NEVER assume it. This leads to the so–called LIMITING REAGENT (or REACTANT) problem. Techniques for dealing with the Limited Reagent (or Reactant) problem are straightforward. (a) Convert all reagent MASSES to MOLES (b) To decide which of the reagents is LIMITING, simply divide the number of moles of each reagent by its coefficient in the BALANCED equation. (c) INSPECT the numbers so obtained; the smallest (or smaller if only two) value indicates the LIMITING REAGENT. The moles of products will depend ONLY on the moles of this reactant – all the others being in EXCESS.
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3 Example : P 4 is made by the following reaction: 2Ca 3 (PO 4 ) 2 (s) + 6SiO 2 (s) + 5C(s) P 4 (s) + 6CaSiO 3 (s) + 5CO 2 (g) Calculate the maximum mass of P 4 (MM =123.9 g/mole) that can be made by reacting the following: Reactants Mass (g) Molecular Mass (g/mol) Ca 3 (PO 4 ) 2 6.00 310. SiO
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Lecture Notes Week 01 - CHEM*1040 WEEK 1 Review Questions:...

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