Ch5_E_SolStoich - SOLUTION STOICHIOMETRY Section 5.9...

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Unformatted text preview: SOLUTION STOICHIOMETRY Section 5.9 GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS STOICHIOMETRY CALCULATIONS Mass zinc Mass HCl Zinc reacts with acids to produce H22gas. If you Zinc reacts with acids to produce H gas. If you gas. 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 Zn, needed to convert the Zn completely? needed to convert the Zn completely? Zinc reacts with acids to produce H2 gas. If you gas. have 10.0 g of Zn, Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Moles zinc Stoichiometric factor Moles HCl Step 1: Write the balanced equation Zn(s) + 2 HCl(aq) --> ZnCl2(aq) + H2(g) Zn(s) HCl( aq) aq) Step 2: Calculate moles of Zn 10.0 g Zn • 1.00 mol Zn = 0.153 mol Zn 65.39 g Zn Concentation HCl Step 3: Use the stoichiometric factor Zinc reacts with acids to produce H22gas. If you Zinc reacts with acids to produce H gas. If you gas. 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 Zn, needed to convert the Zn completely? needed to convert the Zn completely? ACID-BASE REACTIONS ACID-BASE REACTIONS Titrations Titrations H2C2O4(aq) + 2 NaOH(aq) ---> aq) NaOH( aq) acid base Na2C2O4(aq) + 2 H2O(liq) aq) O(liq) Carry out this reaction using a TITRATION. TITRATION. Step 3: Use the stoichiometric factor 0.153 mol Zn • 2 mol HCl = 0.306 mol HCl 1 mol Zn Titration Titration setup setup Buret contains a solution whose concentration is known exactly. Step 4: Calculate volume of HCl req’d 1.00 L 0.306 mol HCl • = 0.122 L HCl 2.50 mol Oxalic acid, Solution of unknown concentration H2C2O4 Page 1 Titration Titration 1. Add solution from the buret. 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 OH --> 5. At equivalence point moles H + = moles OH moles LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately determine its concentration. determine its concentration. 1.065 g of H 2C2O4 1.065 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentraconcentration of the NaOH? NaOH? 1.065 g of H22C2O4(oxalic acid) requires 35.62 1.065 g of H C2O4 ((oxalic acid) requires 35.62 oxalic mL of NaOH for titration to an equivalence mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? point. What is the concentration of theNaOH? NaOH? Step 1: Calculate moles of H 2C2O4 1.065 g • 1 mol = 0.0118 mol 90.04 g Step 2: Calculate moles of NaOH req’d 0.0118 mol acid • 2 mol NaOH = 0.0236 mol NaOH 1 mol acid 1.065 g of H22C2O4(oxalic acid) requires 35.62 1.065 g of H C2O4 ((oxalic acid) requires 35.62 oxalic mL of NaOH for titration to an equivalence mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? point. What is the concentration of theNaOH? NaOH? 1.065 g of H22C2O4(oxalic acid) requires 35.62 1.065 g of H C2O4 ((oxalic acid) requires 35.62 oxalic mL of NaOH for titration to an equivalence mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? point. What is the concentration of theNaOH? NaOH? Step 1: Calculate moles of H2C2O4 = 0.0118 mol acid 0.0118 Step 2: Calculate moles of NaOH req’d LAB PROBLEM #2: LAB PROBLEM #2: Use standardized NaOH to determine Use standardized NaOH 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) ---> aq) NaOH( aq) Na2C4H4O5(aq) + 2 H2O(liq) aq) O(liq) 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? Step 1: Calculate moles of H 2C2O4 = 0.0118 mol acid 0.0118 Step 2: Calculate moles of NaOH req’d = 0.0236 mol NaOH Step 3: Calculate concentration of NaOH = 0.0236 mol NaOH Step 3: Calculate concentration of NaOH 0.0236 mol NaOH = 0.663 M 0.03562 L [NaOH] = 0.663 M NaOH] Page 2 76.80 g of apple requires 34.56 mL of 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is 0.663 M NaOH for titration. What is weight % of malic acid? weight % of malic acid? Step 1: M•V = = Step 2: Calculate quantity of NaOH used. (0.663 M)(0.03456 L) 0.0229 mol NaOH Calculate quantity of acid titrated. titrated. 1 mol acid 2 mol NaOH 76.80 g of apple requires 34.56 mL of 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is 0.663 M NaOH for titration. What is weight % of malic acid? weight % of malic acid? Step 1: = Step 2: = Calculate moles of NaOH used. 0.0229 mol NaOH Calculate moles of acid titrated 0.0115 mol acid 134 g = 1 .54 g mol 76.80 g of apple requires 34.56 mL of 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is 0.663 M NaOH for titration. What is weight % of malic acid? weight % of malic acid? Step 1: = Step 2: = Step 3: = Calculate moles of NaOH used. 0.0229 mol NaOH Calculate moles of acid titrated 0.0115 mol acid Calculate mass of acid titrated. titrated. 1.54 g acid 1.54 g • 1 00% = 2.01% 76.80 g 0.0229 mol NaOH • Step 3: Calculate mass of acid titrated. titrated. 0.0115 mol acid • = 0.0115 mol acid Step 4: Calculate % malic acid. Page 3 ...
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