__NACl___ The NaHCO3 is the limiting reactant and the HCl is the excess reactant in this experiment. Determine the theoretical yield of the NaCl product‚ showing all of your work in the space below. (5 points) 12.71 NaHCO3 / 84.01 g/mol = 0.1513 moles 0.1513 moles * 58.44 g/mol (NaCL molar mass) = 8.84197 g What is the actual yield of NaCl in your experiment? Show your work below. (4 points) 31.52 g 24.35 g = 7.17 g Determine the percent yield of NaCl in your experiment‚ showing all work neatly in the space
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Purpose The purpose of the experiment is to determine the percent yield of the precipitate; by performing double displacement reaction between solutions of two different compounds. Introduction First of all when making a solution of two different compounds; there will many variables that can be considered during the experiment. However‚ the variables are controlled variables. Controlled Variables ∙amount of water that will be dissolved with the compound (amount of water until the compound is
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Dr. Mbah December 6‚ 2012 Percent Yield of a Chemical Reaction Introduction Yield is the quantity of product in a chemical reaction‚ the theoretical yield of a reaction can be calculated using mole ratios from the balanced chemical reaction. The actual yield has to be obtained and measured in a laboratory. It may be usual to often find the actual yield to be less than the theoretical yield due to many different factors. This gives rise to the concept of percent yield. Sodium bicarbonate and hydrochloric
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The percent yield was calculated to be 171% indicating a source of error. The high percent yield was due to the product being wet since we did not have enough time to dry the product with the vacuum filter—since we had to evacuate the building due to the fire alarm. If we were able to vacuum filter our product more‚ the percent yield would be close to 100%. 1. Assign the peaks in your NMR spectrum of your salicylic acid. See NMR spectrum on back 2. Assign the peaks in the IR spectrum of your salicylic
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The percent yield of meso-1‚2-dibromo-1‚2-diphenylethane was calculated to be 66.36% and with a melting point of 235.1°C. Therefore‚ it can be concluded that this experiment was successfully conducted as the percent yield obtained is only 33.64% off from the equilibrium point‚ and because the melting point met the literature value. The percent yield for the product was less than 100%‚ indicating that were experimental errors‚ such as an undesirable side reaction‚ or more likely‚ an incomplete reaction
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2013 Percent Yield of Calcium Carbonate Introduction: The purpose of this experiment is to examine the percent yield of a precipitate in a double displacement reaction. A solution of calcium citrate and sodium carbonate were mixed together‚ then the products were filtered out as so only the precipitate remained. The filtered paper was then dried and the mass of the precipitate in the experiment divided by the theoretical mass of the precipitate from the calculated gave the percent yield. The
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| Percent Yield Lab | | | | 4/20/2012 | Mrs.Sardella Per4 Matt ‚ Kait Mrs.Sardella Per4 Matt ‚ Kait | Introduction *Limiting Reactant: A reactant that is completely consumed during a chemical reaction‚ limiting the amount of product that is produced. *Excess Reactant: A reactant that remains after a reaction is over. *Theoretical Yield: The amount of product that is predicted by stoichiometric calculations *Actual Yield: The amount of product that is recovered
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Description Whenever we do experiments‚ the actual result is a little different from the result we predicted. In chemistry‚ this discrepancy is compared by calculating the percent yield. In this lesson‚ we will define percent yield‚ and go over a few examples. !!!What is Percent Yield? Sometimes you look at a recipe and it says how many servings it will make if you follow the amounts listed on there. There are times‚ however‚ that the number of servings after you have made the food is not the
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conditions. The experiment resulted in 8.29grams of purified compound‚ which is a 66.27 percent yield. Procedure: As per handout with changes Equation: Mechanism: Results: (Scan IR spectrum) Compounds t-Butyl Chloride(Experimental) t-Butyl Chloride(Book Value) Molecular Weight (g/mol) 92.5681 92.5681 Density (g/cm3) 0.842 0.842 Boiling Point (°C) 48 50 Refractive Index 1.3855 1.3857 Theoretical Yield: Let A = t-Butyl Alcohol Let C = t-Butyl Chloride 10.02gA x 1 mol A x 1 mol C
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The NaHCO3 is the limiting reactant and the HCl is the excess reactant in this experiment. Determine the theoretical yield of the NaCl product‚ showing all of your work in the space below. (5 points) NaHCO3 – The Empty Dish 37.06 (grams) – 24.35 (grams) = 12.71 (grams) NaHCO3 12.71 (grams) NaHCO3 ÷ 84.01 (grams/mole) = 0.1513 moles of NaHCO3 0.135 Moles of NaHCO3 × 58.4428 (Molar mass of NaCl) Giving me 8.8240 (grams) NaCl as my theoretical yield. 4. What is the actual yield of NaCl in your experiment
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