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Unformatted text preview: spano velasco (ds28282) hw07 Fakhreddine (52395) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 1.0 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. 1.66 M correct 2. 0.667 M 3. 3.33 M 4. 1.17 M 5. 0.832 M 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 002 1.0 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. 40 mL 3. 5.0 mL 4. 20 mL correct 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 spano velasco (ds28282) hw07 Fakhreddine (52395) 2 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 3 003 1.0 points 200 mL of 1.5 M HCl is mixed with 150 mL of 3.0 M NaOH. HCl + NaOH NaCl + H 2 O What is the molarity of the resulting salt solution? 1. 0.86 M NaCl correct 2. 2.0 M NaCl 3. 1.5 M NaCl 4. 2.25 M NaCl 5. 0.45 M NaCl Explanation: V 1 = 200 mL M 1 = 1 . 5 M V 2 = 150 mL M 2 = 3 . 0 M Our first step is to determine the number of moles of each reactant present: ? mol HCl = 200 mL soln 1 L soln 1000 mL soln 1 . 5 mol HCl 1 L soln = 0 . 3 mol HCl ? mol NaOH = 150 mL soln 1 L soln 1000 mL soln 3 . 0 mol NaOH 1 L soln = 0 . 45 mol NaOH From the balanced chemical equation, we know that one mole of HCl is needed to re- act with each mole of NaOH. We dont have enough HCl to react all the NaOH, so HCl is...
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This note was uploaded on 04/08/2009 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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solution7_pdf - spano velasco (ds28282) hw07 Fakhreddine...

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