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Chemical Reaction Stoichiometry
The document discusses reaction stoichiometry and percent yield calculations for various chemical reactions. It provides examples including combustion of glucose and the Haber process, detailing the formulas for calculating theoretical yield and percent yield. Additionally, it covers limiting reactants and includes several problem statements for calculating these metrics.
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Chemical Reaction Stoichiometry
- 1. Reaction Stoichiometry Dr. K.Shahzad Baig Memorial University of Newfoundland (MUN) Canada Petrucci, et al. 2011. General Chemistry: Principles and Modern Applications. Pearson Canada Inc., Toronto, Ontario. Tro, N.J. 2010. Principles of Chemistry. : A molecular approach. Pearson Education, Inc.
- 2. Practical Matters inReaction Stoichiometry Percent yield represents the ratio between what is experimentally obtained and what is theoretically calculated, multiplied by 100%. % 𝑦𝑖𝑒𝑙𝑑 = 𝐴𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 𝑥 100 how much water is produced when 12.0 g of glucose (C6H12O6) is burned with enough oxygen. C6H12O6 + 6O2 → 6CO2 + 6H2O 12.0 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 𝑥 1 𝑚𝑜𝑙 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 180.0 𝑔 𝑥 6 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑤𝑎𝑡𝑟 1 𝑚𝑜𝑙 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 𝑥 18.0𝑔 1 𝑚𝑜𝑙 𝑤𝑎𝑡𝑒𝑟 = 7.20 𝑔 % 𝑦𝑖𝑒𝑙𝑑 = 6.50 𝑔 7.20 𝑔 𝑥 100 = 90 %
- 3. Problem statement For thereaction 2S(s) + 3O2(g) →2 SO3(s) 2.0 moles of sulfur are placed in a container with an excess of oxygen and 1.7 moles of SO3 are collected. What is the percentage yield?
- 4. Problem statement The Haberprocess can be used to produce ammonia, NH3, and it is based on the following reaction. N2(g)+3H2(g)→2NH3(g) If one mole each of N2 and H2 are mixed and 0.50 moles of NH3 are produced, what is the percent yield for the reaction? calculate the theoretical yield Given that 0.667 moles will produce 0.50 moles of NH3
- 5. Problem statement What isthe percent yield of the following reaction if 60 grams of CaCO3 is heated to give 15 grams of CaO+ CaCO3 → CaO + CO2 Theor. yield = Now calculate the percent yield.
- 6. Problem statement If thereaction of 77.0 grams of CaCN2 produces 27.1 grams of NH3, what is the percent yield? CaCN2+3H2O → CaCO3+2NH3 We know that CaCN2 and NH3's molar masses are 80 g/mol and 17 g/mol, respectively, therefore
- 7. So, the percentyield, defined as the actual yield divided by the theoretical yield and multiplied by 100%, is
- 8. Problem statement What wouldbe the limiting reagent if 26.0 grams of C3H9N were reacted with 46.3 grams of O2? 4C3H9N + 25O2 => 12CO2 + 18H2O + 4NO2 From C₃H₉N: considering CO2 is a product From O₂: The O₂ gives the smaller amount of CO₂, so the O₂ is the limiting reactant.
- 9. So, the percentyield, defined as the actual yield divided by the theoretical yield and multiplied by 100%, is
- 10. Problem statement At 273.15K and 100kPa, 58.34 g of HCl reacts with 0.35 mol of MnO2 to produce 7.056 dm3 of chlorine gas. Calculate the theoretical yield of chlorine. Which one of the two reactants acts as a limiting reactant more moles of hydrochloric acid are present than need, therefore, Manganese dioxide will act as a limiting reagent,
- 11. Theoretically, to produceone mole of chlorine gas for every mole of manganese dioxide would reacts. to calculate the percent yield of the reaction
Editor's Notes
- #3 a 1:6 mole ratio between glucose and water, you can determine how much water you would get by
- #7 we can see that we have a 1:2 mole ratio between CaCN2 and NH3; that is, for every mole of CaCN2 used in the reaction, 2 moles of NH3 will be formed.
- #8 According to our mole-to-mole ratio, 0.96 moles of CaCN2 should have produced 2⋅0.96=1.92 moles of NH3, but instead produced 1.6 moles. This means that not all CaCN2 reacted <=> water is the limiting reagent.
- #10 According to our mole-to-mole ratio, 0.96 moles of CaCN2 should have produced 2⋅0.96=1.92 moles of NH3, but instead produced 1.6 moles. This means that not all CaCN2 reacted <=> water is the limiting reagent.
- #11 This mole ratio tells you that the reaction will always consume 4 moles of hydrochloric acid for every mole of manganese dioxide that takes part in the reaction.
- #12 Under these conditions for pressure and temperature, one mole of any ideal gas occupies 22.7 L → this is known as the molar volume of a gas at STP.