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# HW8 - momin(rrm497 Homework 8 Sutclie(53120 This print-out...

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momin (rrm497) – Homework 8 – Sutclife – (53120) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore answering. 001 10.0 points 183.3 mL o± an HBr solution reacted exactly with 62.1 mL o± 0.0400 M Ca(OH) 2 solution. Calculate the molarity o± the HBr solution. 1. 0.0136 M 2. 0.0400 M 3. 0.1084 M 4. 0.0271 M correct 5. 0.0542 M 6. 0.0379 M Explanation: V 1 = 183 . 3 mL V 2 = 62 . 1 mL M 2 = 0 . 0400 M The balanced equation ±or the reaction is 2 HBr + Ca(OH) 2 CaBr 2 + 2 H 2 O Molarity is moles solute per liter o± solution. We know the volume o± the HBr solution. I± we could Fnd the moles o± HBr in the solution we could calculate the molarity. We start by calculating the moles o± Ca(OH) 2 present: ? mol Ca(OH) 2 = 62 . 1 mL soln × 1 L soln 1000 mL soln × 0 . 0400 mol Ca(OH) 2 1 L soln = 0 . 00248 mol Ca(OH) 2 Using the mole-to-mole ratio ±rom the bal- anced chemical equation we calculate the moles o± HBr needed to react with this amount o± Ca(OH) 2 : ? mol HBr = 0 . 00248 mol Ca(OH) 2 × 2 mol HBr 1 mol Ca(OH) 2 = 0 . 00497 mol HBr This is the number o± moles o± HBr needed to react the Ca(OH) 2 and there±ore the number o± moles present in the 183.3 mL o± HBr so- lution. We can calculate the molarity o± the HBr solution by dividing the moles o± HBr by the volume o± the HBr solution: ? M HBr = 0 . 00497 mol HBr 0 . 1833 L soln = 0 . 0271 M HBr 002 10.0 points Calculate the reaction enthalpy ±or the com- bustion o± ethanol CH 3 CH 2 OH + 3 O 2 2 CO 2 + 3 H 2 O given the data C 2 H 2 + 5 2 O 2 2 CO 2 + H 2 O Δ H rxn = - 1300 kJ/mol C 2 H 2 + H 2 C 2 H 4 Δ H rxn = - 175 kJ/mol C 2 H 4 + H 2 O CH 3 CH 2 OH Δ H rxn = - 44 kJ/mol H 2 + 1 2 O 2 H 2 O Δ H rxn = - 286 kJ/mol 1. - 1367 kJ/mol correct 2. - 1761 kJ/mol 3. - 1800 kJ/mol 4. None o± these is correct. 5. - 1717 kJ/mol Explanation: 003 10.0 points ²or which reactions I) O 2 (g) + H 2 (g) H 2 O 2 ( ) II) C(s , diamond) + O 2 (g) CO 2 (g) III) N 2 ( ) + 3 ² 2 (g) 2 N² 3 ( ) IV) C(s , graphite) + 3 2 O 2 (g) + H 2 (g) CO 2 (g) + H 2 O(g) V) 2 ²e(s) + 3 2 O 2 (g) ²e 2 O 3 (s)

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momin (rrm497) – Homework 8 – Sutclife – (53120) 2 would Δ H r = Δ H f ? 1. I, II, III, IV and V 2. I and II only 3. I and V only correct 4. II and IV only 5. II, III and IV only 6. IV and V only 7. I, III and V only 8. I, III, IV and V only Explanation: 004 10.0 points Calculate the standard reaction enthalpy For the Formation oF boron tri±uoride, which is widely used in the chemical industry: B 2 O 3 (s) + 3 Ca² 2 (s) 2 B² 3 (g) + 3 CaO(s) . The standard enthalpy oF Formation For B² 3 is - 1137 kJ / mol, For CaO - 635 . 09 kJ / mol, For B 2 O 3 - 1272 . 8 kJ / mol, and For Ca² 2 - 1219 . 6 kJ / mol. 1. 800 . 22 kJ / mol 2. 892 . 4 kJ / mol 3. 650 . 03 kJ / mol 4. 832 . 45 kJ / mol 5. 752 . 33 kJ / mol correct 6. 550 . 2 kJ / mol Explanation: Δ H f , BF 3 (g) = - 1137 kJ / mol Δ H f , CaO(s) = - 635 . 09 kJ / mol Δ H f , B 2 O 3 (s) = - 1272 . 8 kJ / mol Δ H f , CaF 2 (s) = - 1219 . 6 kJ / mol B 2 O 3 (s) + 3 Ca² 2 (s) 2 B² 3 (g) + 3 CaO(s) Using Hess’ Law, Δ H rxn = p s n Δ H j P products - p s n Δ H j P reactants = 2 Δ H f , BF 3 (g) + 3 Δ H f , CaO(s) - b Δ H f , B 2 O 3 (s) + Δ H f , CaF 2 (s) B = 2 ( - 1137 kJ / mol) + 3 ( - 635 . 09 kJ / mol) - b - 1272 . 8 kJ / mol + 3 ( - 1219 . 6 kJ / mol) B = 752 . 33 kJ / mol .
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HW8 - momin(rrm497 Homework 8 Sutclie(53120 This print-out...

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