Question

i uploaded everything that was given. Where would I find the experiemental values?

Complete the calculation summary below. READ THE EXPERIMENTAL DISCUSSION FIRST! REPORT ALL ANSWERS TO THE CORRECT SIGNIFICANT FIGURES. SHOW CALCULATION SET-UP on the next page. PART B: Molarity Determination via Solution Stoichiometry 7.625 g Mass of Empty 50 mL Beaker 7.976 g Mass of Beaker and NaCl (final heating) Mass of NaCl Produced (Experimental Yield of NaCl) moles Moles of Na,CO, Consumed 0.00500 Liter Volume of Na,CO, Solution mole/Liter Molarity of Na,CO, Solution PART C: Reaction Stoichiometry 8.125 g Mass of Empty 50 mL Beaker 8.387 g Mass of Beaker and Unknown Solid Mass of Unknown Solid 8.405 g Mass of Beaker and NaCl (final heating) Mass of NaCl Produced (Experimental Yield of NaCI) Theoretical Yield of NaCl if Unknown Solid is Na,cO3 grams NaCl Theoretical Yield of NaCl if Unknown Solid is NaHCO grams NaCl Formula of the Unknown Solid Percent Yield Calculation Set-Ups: Advance Problem Assignment 105.986 FORMULA MASS: NaCl 58.441 NaHCO, =84.005 Na,CO, EXPERIMENT 5 Reaction Stoichiometry: Mass Relationships in Balanced Chemical Equations According to the principle of conservation of mass, the total mass of each element in the product of a chemical reaction must be identical to the toeal mass of each element in the reactant. For this reason chemical equations, our descriptions of chemical change, must be balanced utilizing appropriate numerical coefficients. The balanced equation specifies the relationships between moles/mass of reactants consumed and moles/mass of products formed. These relationships dict amounts of reactant consumed and/or product formed. In some situations the stoichi- ometry can also help identify a particular reactant or product. are referred to as reaction stoichiometry" and are used to pre- For instance, iron forms two different chloride compounds, FeCl, and FeCl,. Boch of these react in aqueous solution with magnesium metal, forming elemental iron and magnesium chloride: Fe MgCl Mg FeCl) 1. +2 Fe 3MgCl 2FECI 3Mg 2. Note that one mole of iron(III) chloride produces one and a half times the amount of mag- nesium chloride that would be produced by one mole of iron(II) chloride. If one experi- mass of iron measures the amount of magnesium chloride produced by a specific mentally chloride, the identity of the iron compound can be determined by comparison with the stoichiometric mass of magnesium chloride that is predicted by each reaction. A 1.00 gram sample of iron(II) chloride (FeCl,) will theoretically produce 0.751 grams of MgCI FeCI) (95.218 g/1 mol MgCI,] (1.00 g FeCl,) 1 mol/126.75 g] [1 mol MgCl,/1 mol = 0.751, g However, 1.00 gram of iron(III) chloride (FeCl) will produce 0.881 grams of MgCL: (1.00 g FeCl,)1 mol/162.206 g] [3 mole MgCI,/2 mol FeCl,] [95.218 g/1 mol MgCl,] 0.880, g If the experimentally measured mass of magnesium chloride actually produced was 0.868 conclude that the reacting compound was iron(II) chloride. On the other grams, we can hand, if the actual mass of MgCl, were close to 0.751 grams, iron(II) chloride would be identified as the reacting compound. Many reactions occur berween solutes in solution phase. One often knows what the solute is but not its concentration (molariry). The stoichiometry of the reaction can be urilized to determine concen tration if the formala and the equarion is known. For cxample, experimental measurement of the mass of magnesium chloride produced by an aqueous solution sample of FeCl, according to reaction 2 allows one to calculate the moles of FeCI, solute presene in the sample (2 mol FeCl, for every 3 mole of MgCI, produced). Concentration of the reactant in moles per liter can then be determined from the calculated moles of FeCl, solute and the volume of the FeCl, solution sample used in the reaction tf, for example, a 10.00 mL sample of a FeCI, solution produced 0.440 g of MgCl, upon reaction with magnesium, the molarity of that solution is determined to be 0.308, M via equation 2: (0.440 g)(1 mol MgCl,/95.218 g][2 mol FeCl,/3 mol MgCI,] 3.08, x 10 mol FeCI, Molarity 3.08, x 10- moles FeCl,/0.01000 liters of solution = 0.308, M The reactions of sodium carbonate and sodium hydrogen carbonate with hydrochloric acid both produce carbon dioxide gas, sodium chloride, and water. Heating easily removes water vapor and the gas, allowing the yield of sodium chloride to be readily determined by mass measurement. The balanced equations are 3. NaHCO, но но NaCl нC 2 Hа COse + 4. Na,COy) 2 NaCl CO%e + Note that only one mole of NaCl is produced per mole of NaHCO, but two moles of NaCl are pro- duced per mole of Na,CO. Therefore, if one knows the mass but not the formula of the compound reacting with HCl, and one can determine the mass of NaCI produced as a result (the Experimental Yield of NaCl), the reactant can be identified based on the equation stoichiometry in a manner simi- lar to the iron chloride examples (reactions 1 and 2) discussed above. This experiment will employ reactions 3 and 4 above to demonstrate the concepts described for the iron chloride reactions because the product is easily purified by evaporation. The mass of sodium chloride produced is related to the nature of the reacting compound by the stoichiometry of reactions 3 and 4 in the same manner as the mass of magnesium chloride produced in reactions 1 and 2 was related to the nature of the iron chloride compounds. TERMS AND PROCEDURES RELATED TO THIS EXPERIMENT The unknown solution container is fitted with a penny head stopper. Grip the stopper between the first and second fingers of your hand with your palm facing up. Withdraw the stopper, turn your hand so the stopper is pointing away from you, and grasp the bottle with the SAME hand using the little finger to support the bottom of the bottle. Hold the receiving container in your other hand and pour the required volume of solution into the receiving container. The desired volume is APPROXIMATE. Do not be concerned if your actual volume is a little more or less than the amount specified in the laboratory procedure. Catch the drip from the bottle with the edge of the receiving container, put the bottle down on the bench, and insert the stopper. Put the container down and RINSE THE CLOSED BOTTLE under the tap water in preparation for the next student. NOTE that the stopper is NEVER allowed to contact the laboratory bench. Ask your instructor to demonstrate this technique for you if you are uncertain about these instructions. EXPERIMENT S WEAR YOUR SAFETY GOGGLES1/LAB COAT! WASH YOUR HANDS WHEN FINISHED! EXPERIMENTAL PROCEDURE PART A: Preparation In this experiment, two clean and dry 50 mL beakers will be used as reaction vessels. The white "marking area" should be clean to permit identification of each reaction. Set up two small test tubes with Pasteur pipet droppers in a test tube rack. Put about 50 drops (2.5 mL-about 2 cm or 3/4 inch depth in a small test tube) of distilled water in one and 50 drops of 6 M hydrochloric acid HCl in the ocher. LABEL both test tubes. Electric hoe plates (heat setting # 2.5), for use in the evaporation step, are set up for you in the laboratory hood arca. PART B: Molarity Determination Via Solution Stoichiometry Please review "APPENDIX: Use of a Volumetric Pipet" before you begin the experiment. Do NOT pipet by mouth suction. Wear safety goggles and lab coat and/or apron. Use a pencil to mark one mination); weigh and record the mass 50 mL beaker with your initials and the letter "M" (for MOLARITY deter- to the nearest milligram. Obtain a pipet pump and a 5.00 mL. volumetric pipet from the laboratory cart. Use your 10 mL graduated cylinder to measure about 7 to 9 mL. of the unknown sodium carbonate solution assigned by your instructor. Then use the volumetric pipet to extract 5.00 mL of this solu- tion from the cylinder and deliver it into the weighed beaker. Record the unknown solution number. Set the excess unknown solution aside for later use in PART D. Add 6 M HCl drop wise (Eye Protection!) until effervescence (bubbles) ceases. Add five drops at a time and then GENTLY swirl the container. As the gaseous product, carbon dioxide, escapes, you'll observe effervescence. Continue adding HCl in five-drop increments, mixing in between, until there is no sign of reaction (this usually requires 20 to 30 drops of acid). Then put the beaker on a hot plate in the hood and allow the liquid to evaporate. Go on to PART C while the evaporation takes place. on PART C. Occasionally check the evaporation process while working Assuming the unknown solid was sodium hydrogen carbonate; calculate the theoretical yield of sodium chloride that should have been produced by your sample. Compare these theoretical values with your experimental yield of NaCL Which solid was your un- known, NaHCO, or Na,CO,? Once you have identified the solid, calculate PERCENT YIELD for your reaction. PART D: Clean Up Eye Protection! The solid in the beakers is sodium chloride (table salt.) Add approximately 20 mL of tap water to each beaker, stir, combine the solutions and place the beaker in a sink. You also have 10 or 20 drops of hydrochloric acid and 2-3 mL of Na,CO, solution left rinse their containers, and add the rinse to the same beaker. (You should observe some effervescence as the sodium carbonate neutralizes the excess hydrochloric acid.) over. Pour both of these into the beaker, Add 4-6 drops of phenolphthalein indicator to the mixture in the beaker. If the mixture remains clear (water white) in color when phenolphthalein is introduced, add 6.0 M NaOH one drop at a time, with stirring, until one drop causes the solution to turn pink. Then add 1.0M HC to fade clear (water white). CAUTION: Do not add excess HCL drop at a time, with stirring, until one drop causes the pink color one If the mixture turns pink in color, add 6.0 M HCl one drop at a time, with stirring, until one drop causes the pink color to fade clear (water white). pink indicating the Then add one drop of 6.0 M NaOH-the solution should again solution is now turn drop basic. one Then add 1.0 M HCl one drop at a time, with stirring, until one drop causes the pink color to fade clear (water white). CAUTION; Do not add excess HCL O Rinse the neutralized solution into the sink drain with flowing tap water. Rinse the pipet with tap water and return the pipet and pipet pump to the laboratory cart. Rinse the HCl from the Pasteur pipet dropper with tap water and discard the pipet into the GLASS WASTE container provided in the laboratory.

Answer 1

part B

mass of NaCl produced = 7.976 - 7.625 = 0.351 gram

-----------------

we know the reaction of preparation of NaCl from Na2CO3 is Na2CO3 + 2 HCl -----> H2CO3 + 2NaCl

from this reaction we get to know that

105.986 gm of Na2CO3 produces = 2x 58.442 gram of NaCl hence 1 gm NaCl needs = 109.986 / 116.882 = 0.907 gm

hence 0.351 gm NaCl produced by = 0.907 x 0.351 = 0.318 gm

now we know 1 mole of Na2CO3 = 105.986 gm

or we can say 105.986 gm of Na2CO3 = 1 mole

this says 1 gm = 1/105.986 gm

hence 0.318 gm = [1 / 105.986] x 0.318 = 0.003 moles of Na2CO3

--------------

​​​​​​molarity of Na2CO3 = moles of Na2CO3 / volume in liter = 0.003 moles / 0.005 liter = 0.6 M is molarity

--------part C -----

mass of unknown solid = mass of beaker with unknown - mass of beaker only

= 8.387 - 8.125 = 0.262 gram

----------

mass of NaCl produced = ,mass of beaker with NaCl - mass of beaker only

= 8.405 - 8.125 =   0.280 grams

--------------

theoretical yield of NaCl will be calculated by eqation

Na2CO3 + 2 HCl -----> H2CO3 + 2 NaCl given that unknown solid is Na2CO3 [ 0.262 gram]

105.986 gram Na2CO3 produces = 2x58.441 = 116.882 gram NaCl

hence 0.280 gram of Na2CO3 will produce = [116.882/105.986]x0.262 = 0.289 grams of NaCl theoretically

--------------

if unknown solid is NaHCO3

the eqations belongs to it

NaHCO3 + HCl -----> NaCl + H2CO3 this eqation says 84.005 gram of NaHCO3 produces = 58.441 grams of NaCl

hence 0.262 gram of NaHCO3 will produce = [58.441/84.005]x0.262 = 0.182 gram of NaCl produced

----------