Homework 08 - ycs73 Homework 8 Fakhreddine (53605) 1 This...

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Unformatted text preview: ycs73 Homework 8 Fakhreddine (53605) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points 500 mL of 2 M sodium hydroxide is mixed with 200 mL of 2 M sulfuric acid. What is in this reaction vessel after the reaction has run to completion? 1. water, SO 2- 4 , NaOH molecules, Na + , and OH 1- 2. water, Na 2 SO 4 molecules, and NaOH molecules 3. water, SO 2- 4 , Na + , and OH- correct 4. water, Na + , H 2 SO 4 molecules, H 3 O + , and SO 2- 4 5. water, Na 2 SO 4 molecules, and H 2 SO 4 molecules 6. water, SO 2- 4 , Na + , and NaOH molecules 7. water, SO 2- 4 , Na + , and H 3 O + Explanation: 002 10.0 points We find that 61.1 mL of an HCl solution reacts exactly with 74.2 mL of 1.371 M KOH solution. Calculate the molarity of the HCl solution. 1. 1.17 M 2. 0.832 M 3. 3.33 M 4. 0.667 M 5. 1.66 M correct Explanation: V 1 = 61 . 1 mL V 2 = 74 . 2 mL M 2 = 1 . 371 M The balanced equation for the reaction is HCl + KOH KCl + H 2 O Molarity is moles solute per liter of solution. We know the volume of the HCl solution. If we could find the moles of HCl in the solution we could calculate the molarity. We start by calculating the moles of KOH present: ? mol KOH = 74 . 2 mL soln 1 L soln 1000 mL soln 1 . 371 mol KOH 1 L soln = 0 . 1017 mol KOH Using the mole-to-mole ratio from the bal- anced chemical equation we calculate the moles of HCl needed to react with this amount of KOH: ? mol HCl = 0 . 1017 mol KOH 1 mol HCl 1 mol KOH = 0 . 1017 mol HCl This is the number of moles of HCl needed to react with the KOH and therefore the num- ber of moles present in the 61.1 mL of HCl solution. We can calculate the molarity of the HCl solution by dividing the moles of HCl by the volume of the HCl solution: ? M HCl = . 1017 mol HCl . 0611 L soln = 1 . 66 M HCl 003 10.0 points How many milliliters of 0.010 M HNO 3 will neutralize 20 mL of 0.0050 M Ba(OH) 2 ? 1. 10 mL 2. 20 mL correct 3. 40 mL 4. 5.0 mL ycs73 Homework 8 Fakhreddine (53605) 2 Explanation: [HNO 3 ] = 0.010 M V Ba(OH) 2 = 20 mL [Ba(OH) 2 ] = 0.0050 M The balanced equation for this neutraliza- tion reaction is 2 HNO 3 + Ba(OH) 2 Ba(NO 3 ) 2 + 2 H 2 O We determine the moles of Ba(OH) 2 present: ? mol Ba(OH) 2 = 0 . 020 L soln . 0050 mol Ba(OH) 2 1 L soln = 0 . 00010 mol Ba(OH) 2 Using the mole ratio from the balanced chemical equation we calculate the moles of HNO 3 needed to react with this amount of Ba(OH) 2 : ? mol HNO 3 = 0 . 00010 mol Ba(OH) 2 2 mol HNO 3 1 mol Ba(OH) 2 = 0 . 00020 mol HNO 3 We use the molarity of HNO 3 solution to convert from moles to volume of HNO 3 : ? mL HNO 3 = 0 . 00020 mol HNO 3 1000 mL soln . 010 mol HNO 3 = 20 mL HNO 3 004 10.0 points 200 mL of 1.5 M HCl is mixed with 150 mL of 3.0 M NaOH....
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Homework 08 - ycs73 Homework 8 Fakhreddine (53605) 1 This...

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