Ch5_E_SolStoich

# Ch5_E_SolStoich - SOLUTION STOICHIOMETRY Section 5.9 Zinc...

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Unformatted text preview: SOLUTION STOICHIOMETRY Section 5.9 Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Zinc reacts with acids to produce H22gas. If you Zinc reacts with acids to produce H gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? needed to convert the Zn completely? Step 3: Use the stoichiometric factor Step Use 0.153 mol Zn • 2 mol HCl = 0.306 mol HCl 1 mol Zn GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS STOICHIOMETRY CALCULATIONS Mass HCl Mass zinc Moles zinc Stoichiometric factor Moles HCl Zinc reacts with acids to produce H22gas. If you Zinc reacts with acids to produce H gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? needed to convert the Zn completely? Step 1: Write the balanced equation Step Write Zn(s) + 2 HCl(aq) --> ZnCl2(aq) + H2(g) Step 2: Calculate moles of Zn Step Zn 10.0 g Zn • Concentation HCl ACID-BASE REACTIONS ACID-BASE REACTIONS Titrations Titrations H2C2O4(aq) + 2 NaOH(aq) ---> acid base acid base Na2C2O4(aq) + 2 H2O(liq) Carry out this reaction using a TITRATION . Carry TITRATION Step 4: Calculate volume of HCl req’d Step Calculate 1.00 L 0.306 mol HCl • = 0.122 L HCl 2.50 mol Oxalic acid, H2C2O4 Page 1 1.00 mol Zn = 0.153 mol Zn 65.39 g Zn Step 3: Use the stoichiometric factor Step Use Titration Titration setup setup Buret contains a solution whose concentration is known exactly. Solution of unknown concentration Titration Titration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3. Indicator shows when exact stoichiometric reaction has occurred. 4. Net ionic equation H+ + OH- --> H2O 5. At equivalence point moles H + = moles OH - LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — ii.e., accurately solution of NaOH — .e., accurately determine its concentration. determine its concentration. Step 1: Calculate moles of H 2C2O4 Step Calculate 1.065 g of H 2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? 1.065 g of H22C2O4((oxalic acid) requires 35.62 1.065 g of H C2O4 oxalic acid) requires 35.62 mL of NaOH ffortitration to an equivalence mL of NaOH or titration to an equivalence point. What is the concentration of the NaOH? point. What is the concentration of the NaOH? 1.065 g of H22C2O4((oxalic acid) requires 35.62 1.065 g of H C2O4 oxalic acid) requires 35.62 mL of NaOH ffortitration to an equivalence mL of NaOH or titration to an equivalence point. What is the concentration of the NaOH? point. What is the concentration of the NaOH? Step 1: Calculate moles of H 2C2O4 Step Calculate Step 1: Calculate moles of H2C2O4 Step Calculate = 0.0118 mol acid Step 2: Calculate moles of NaOH req’d Step Calculate = 0.0236 mol NaOH Step 3: Calculate concentration of NaOH Step Calculate 1.065 g of H22C2O4((oxalic acid) requires 35.62 1.065 g of H C2O4 oxalic acid) requires 35.62 mL of NaOH ffortitration to an equivalence mL of NaOH or titration to an equivalence point. What is the concentration of the NaOH? point. What is the concentration of the NaOH? = 0.0118 mol acid Step 2: Calculate moles of NaOH req’d Step Calculate = 0.0236 mol NaOH Step 3: Calculate concentration of NaOH Step NaOH 0.0236 mol NaOH = 0.663 M 0.03562 L [NaOH] = 0.663 M Page 2 1.065 g • 1 mol = 0.0118 mol 90.04 g Step 2: Calculate moles of NaOH req’d Step Calculate 0.0118 mol acid • 2 mol NaOH = 0.0236 mol NaOH 1 mol acid LAB PROBLEM #2: LAB PROBLEM #2: Use standardized NaOH to determine Use standardized NaOH to the amount of an acid in an unknown. the amount of an acid in an unknown. Apples contain malic acid, C4H6O5. C4H6O5(aq) + 2 NaOH(aq) ---> Na2C4H4O5(aq) + 2 H2O(liq) 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? 76.80 g of apple requires 34.56 mL of 76.80 g of apple requires 34.56 mL of 0.663 M NaOH ffor titration. What is 0.663 M NaOH or titration. What is weight % of malic acid? weight % of malic acid? 76.80 g of apple requires 34.56 mL of 76.80 g of apple requires 34.56 mL of 0.663 M NaOH ffor titration. What is 0.663 M NaOH or titration. What is weight % of malic acid? weight % of malic acid? 76.80 g of apple requires 34.56 mL of 76.80 g of apple requires 34.56 mL of 0.663 M NaOH ffor titration. What is 0.663 M NaOH or titration. What is weight % of malic acid? weight % of malic acid? Step 1: Step M•V = = Step 2: Step Step 1: Step = Step 2: Step = Step 1: Step = Step 2: Step = Step 3: Step = Calculate quantity of NaOH used. Calculate (0.663 M)(0.03456 L) 0.0229 mol NaOH Calculate quantity of acid titrated. Calculate 0.0229 mol NaOH • 1 mol acid 2 mol NaOH = 0.0115 mol acid Calculate moles of NaOH used. Calculate 0.0229 mol NaOH Calculate moles of acid titrated Calculate 0.0115 mol acid Step 3: Calculate mass of acid titrated. Step Calculate 0.0115 mol acid • 134 g = 1 .54 g mol Calculate moles of NaOH used. Calculate 0.0229 mol NaOH Calculate moles of acid titrated Calculate 0.0115 mol acid Calculate mass of acid titrated. Calculate 1.54 g acid Step 4: Calculate % malic acid. Step Calculate 1.54 g • 1 00% = 2.01% 76.80 g Page 3 ...
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## This note was uploaded on 10/19/2011 for the course CHM 2210 taught by Professor Staff during the Spring '08 term at University of Central Florida.

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