I. Introduction The Percent Yield lab is designed to further the students’ understanding of percent yield by having them predict how much material will be produced from a reaction; specifically a double replacement reaction. In order to perform the lab‚ a solid understanding of percent yield is necessary. According to Prentice Hall Chemistry book‚ percent yield is comprised of two main components. The first is the theoretical yield. The theoretical yield is what is calculated and predicted. It is
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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) 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 below. (5 points)
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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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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? Show your work below. (4 points) The actual yield is: 31.52 (grams) – 24.35 (grams)
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Yield of CuCl2.2DMSO Formula weight (Mr) of CuCl2 = 63.55 + (35.45 x 2) =134.45g/mol Formula weight of product CuCl2.2DMSO = 134.45 + 2[16 + 32.06 + (12.01 x 2) + (1.0079 x 6)] = 290.704g/mol Mass of CuCl2= 0.850g Equation for reaction CuCl2 + 2DMSO -> CuCl22DMSO Mole ratio between CuCl2 and CuCl22DMSO = 1:1 Mole of CuCl2 = Mass/ Mr = 0.850/134.45 = 0.00632 moles Since the ratio between CuCl2 and CuCl22DMSO = 1:1‚ mole of CuCl2DMSO is also 0.0063 moles. To find theoretical yield of CuCl2
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What are Yield to Maturity (YTM) and Yield to Call (YTC)? By calculating the present and future value of bonds‚ managers can make sound decisions about their potential strengths and weaknesses as investments. Answer the following questions in this week’s Discussion 2 thread: 1. What terms (or inputs) are needed to calculate yield to maturity (YTM)? How does this compare to calculating yield to call (YTC)? To calculate the YTM you will need to use Annual Interest‚ Par value‚ Market Price
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|Chapter 4. Ch04 P24 Build a Model | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |Rework Problem 4-12 using a spreadsheet. After completing questions a through d‚ answer the new question. A 10-year 12 percent semiannual coupon bond‚ with a par value of $1‚000‚ may be called in 4 years at a call price of $1‚060. The bond sells for $1‚100. (Assume that the bond has just been issued.) | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
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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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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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