CHM1045 Lecture 10 Outline of Last LectureI. Chemicali. Chemical reaction definitionii. Chemical equation definition II. How to read Chemical EquationsIII. Balancing Chemical Equationsi. Step 1ii. Step 2iii. Step 3iv. Step 4v. Step 5vi. Example 1Outline of Current LectureI. Amount of Reactants and Productsi. Example 1ii. Example 2II. Limiting Reagenti. Example 3ii. Example 4III. Reaction Yieldi. Example 5Current LectureAmounts of Reactants and Products:Mass (g) of compound A Mass of compound B (use molar mass(g/mol) (use a mole ratio of A&B (Use molar mass of of compound A) from balanced equation) compound B)Moles of compound A Moles of compound B1. Write balance Equation2. Convert Quantities of known substances into moles3. Use coefficients in balanced equations to calculate the # of moles of the sought quanitity4. Convert moles of sought quantity into desired unitsExample 1:The food we eat is degraded, or broken down, in our bodies to provide energy for growth and function. A general overall equation for this very complex process represents the degradation of glucose (C6H12O6) to carbon dioxide (CO2) and water (H2O): If 856 g of C6H12O6 is consumed by a person over a certain period, what is the mass of CO2 produced?Answer:856 C6H12O6 X 1 mol C6H12O6 = 4.750 mol C6H12O6 180.2g C6H12O64.750 molC6H12O6 X 6 mol CO2 = 28.50 mol CO2 1 mol C6H12O628.50 mol cO2 X 44.01g CO2 = 1.25 *10 ^3 g CO2 1 mol CO2Example 2: All alkali metals react with water to produce hydrogen gas and the corresponding alkali metal hydroxide. A typical reaction is that between lithium and water:How many grams of Li are needed to produce 9.89 g of H2?9.89g H2 X 1 mol H2 X 2 mol Li X 6.941g Li = 68.1g Li 2.016 gH2 1 mol H2 1 mol LiLimiting Reagent: - Reactants used up first in the reaction2NO + O2 2NO2NO is the limiting reagentO2 is the excess reagentExample 3:Urea [(NH2)2CO] is prepared by reacting ammonia with carbon dioxide:In one process, 637.2 g of NH3 are treated with 1142 g of CO2. (a) Which of the two reactants is the limiting reagent? (b) Calculate the mass of (NH2)2CO formed.(c) How much excess reagent (in grams) is left at the end of the reaction? a.) moles of (NH2)2CO= 637.2g NH3 1 mol NH3 X 1 mol(NH2)2CO 17.03 g NH3 2 mol NH3= 18.71 mol (NH2)2COmoles of (NH2)2CO= 1142gCO2 X 1 mol CO2 X 1 mol (NH2)2CO 44.01 g CO2 1 mol CO2= 25.59 mol(NH2)CO2NH3 must be the limiting reagent because it produces a smaller amount of (NH2)2COb.) mass of (NH2)2CO= 18.71 mol (NH2)2CO x 60.06g (NH2)2CO 1 mol (NH2)2CO= 1124g (NH2)2COc.) mass of CO2 reacted= 18.71 mol(NH2)2CO x 1 mol CO2 x 44.01gCO2 1 mol (NH2)2CO 1 mol CO2= 823.4 g CO2The amount of CO2 remaing is the difference between the initial amount (1142g) and the amount reacted ( 823.4g) Mass of CO2 remaining= 1142g – 823.4g = 319gExample 4:The reaction between alcohols and halogen compounds to form ethers is important in organic chemistry, as illustrated here for the reaction between methanol (CH3OH) and methyl bromide (CH3Br) to form dimethylether (CH3OCH3), which is a useful precursor to other organic compounds and an aerosol propellant.This reaction is carried out in a dry (water-free) organic solvent, and the butyl lithium (LiC4H9) serves to remove a hydrogen ion from CH3OH. Butyl lithium will also react with any residual water in the solvent, so the reaction is typically carried out with 2.5 molar equivalents of that reagent. How many grams of CH3Br and LiC4H9 will be needed to carry out the preceding reaction with 10.0 g of CH3OH?Grams of CH3Br= 10.00gCH3OH x 1 mol CH3OH x 1 mol CH3Br x 94.93g CH3Br 32.04g CH3OH 1 molCH3OH 1 mol CH3Br= 29.6g CH3Br Grams of Li C4H9= 10.00gCH3OH x 1 mol CH3OH x 2.5 mol LiC4H9 x 64.05g LiC4H9 32.04g CH3OH 1 molCH3OH 1 mol LiC4H9= 50.0g LiC4H9Reaction Yield:Theoretical Yield - is the amount of product that would result if all the limiting reagent reacted.Actual Yield- is the amount of product actually obtained from a reaction.% Yield= Actual Yield X 100% Theoretical YieldExample 5:Titanium is a strong, lightweight, corrosion-resistant metal that is used in rockets, aircraft, jet engines, and bicycle frames. It is prepared by the reaction of titanium(IV) chloride with molten magnesium between 950°C and 1150°C:In a certain industrial operation 3.54 × 107 g of TiCl4 are reacted with 1.13 × 107 g of Mg. (a) Calculate the theoretical yield of Ti in grams. (b) Calculate the percent yield if 7.91 × 106 g of Ti are actually obtained.A.) Moles of TI= 3.54 x10^7g TiCl4 X 1 mol TiCl4 X 1 mol Ti 189.7 g TiCl4 1 mol TiCl4= 1.87*10^5 mol Timoles of Ti= 1.13 *10^7 gMg X 1 mol Mg X 1 mol Ti 23.31 gMg 2 mol Mg= 2.32*10^5 mol Ti1.87*10^5 mol Ti X 47.88g Ti = 8.95 *10^6g Ti1 mol TiB.) % yield = actual yield X 100 % Theoretical yield = 7.91 *10^6 X 100% 8.95
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