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chem lab #3

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    University of Technology General Chemistry Lab 1 (CHY2022) Experiment #3: Determination of mole ratios of chemical reactions  Name: Davion Tracey ID#: 1402673 Course of Study: B. Eng. Mechanical Engineering Lecturer: Mrs Barnett Lab Session: Mondays 11  –   2 pm Lab Date: January 28, 2019 Date of Submission: February 04, 2019    Data Analysis: 1.   Determine the whole number mole ratio of the two reactants . From the graph obtained between Temp Change vs. Volume of Hypochlorite the mole ratio of Hypochlorite to thiosulphate ions was found to be 4:1 2.   The molarities of the reactant solutions were equal in this experiment. Is this necessary, or even important, for the success of the experiment? The aim of the experiment was to determine the mole ratios of the reactants for the reaction. For this experiment the molarities of both NaOCl and the Na 2 S 2 O 3 were known to be 0.50 M, hence the mole ratios of the reactants were obtained comparing the volume of each reactant used. However if the molarities were different the mole ratios could still  be determined by exploiting the formula (X = M x V), where X is the number of moles, Temp Change = mx + bm (slope) = 0.4386 ℃/mL y(intercept) = -0.3845 Temp Change = mx + bM(slope) = -2.0842 ℃/mL y(intercept) = 99.726 Volume of Naocl (40mL) resulting in max temp change051015202505101520253035404550    T   e   m   p   e   r   a   t   u   r   e   C    h   a   n   g   e    (     ℃    ) Volume of Hypochlorite (mL) Temperature change vs. Volume of Hypochlorite  M is the given molar concentration of the reactant and V is the volume of the reactants used. Therefore the Molarities of both reactants didn’t  have to be the same for the success of the experiment. 3.   Identify the limiting factor in each trial? Limiting Reactant Sample Calculation: Using volume of both OCl -  and S 2 O 32-  as 25ml 4 OCl -  + 1 S 2 O 32-  >>>>> Products M of NaOCl = 0.50 M V of NaOCl = 0.025L # Of Moles of available for reaction = M x V = 0.50 x 0.025 = 0.0125 moles  # of moles of Na 2 S 2 O 32-  required for reaction = (0.0125 moles)/4 = 0.003125  moles (because mole ratio is 4:1) # of moles of Na 2 S 2 O 32 available for reaction = M x V = 0.50 x 0.025 = 0.0125 moles  Excess moles of Na 2 S 2 O 32  = 0.0125  –   0.003125 = 0.009375 moles   Nb . Because Na 2 S 2 O 32  is in excess of 0.009375 moles then the limiting reactant is  NaOCl -  because it is not in excess and so will be used up during the reaction therefore  bringing the reaction to an end. Volume OCl - Volume S 2 O 32- Temp Change ( ℃ ) Limiting Reactant 10 40 3.5 OCl - 17 33 6.5 OCl - 25 25 12.5 OCl - 33 17 14 OCl - 40 10 16.4 OCl -  / S 2 O 32- 43 7 10 S 2 O 32-  45 5 6 S 2 O 32-    Exercise: The actual balance equation for the reaction between OCl -  and S 2 O 32  is: 4 OCl -  + S 2 O 32-  + 2 OH -  = 2 SO 42-  + 4 Cl -  + H 2 O Mole ratios calculated matched the actual mole ratio of the reaction.

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