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Chemical Equations

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Published in: Chemistry
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It provide chemical Equation

Sachin S / Pune

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Qualification: M.Sc (Pune University - 2008), B.Sc (Yashwant College - 2006)

Teaches: Chemistry, AIEEE, BITSAT, CET, IIT JEE Advanced, IIT JEE Mains, AIPMT, Medical Entrance Exams, NEET

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  1. Chapter 10 Chemical Calculations and Chemical Equations An Introduction to Chemistry by Prof Sachin Sabne Contact :8888483015
  2. Chapter Map Balance chemical equations (Section 4. l) Convert between names and formulas (Section 5.3) Unit conversio using unit analysis & 8.5 nding an swers Section 8.2 Use molar mass to convert between mass and moles of substance (Sections 9.2 & 9.3) Determining molar mass (Sections 9.2 & 9.3) Write chem iCal equations for precl I cation reactions Section 4.2) nVert between mass of one substance in a chemical reaction and mass of another chemical equations for acid-base reactions (Section 5.6) Percent yie calculations Limiting reactant problems uarion sroichiometry calculations usin Fari molar mass and mo Use molariry as a conversion factor to convert between volume Of solution and moles of solute Identify solute, solvent, and solution (Section 4.2)
  3. Phosphoric Acid-±—.-— • Furnace Process for making H3P04 to be used to make fertilizers, detergents, and pharmaceuticals. — React phosphate rock with sand and coke at 2000 oc. 6Si02 + IOC — React phosphorus with oxygen to get tetraphosphorus decoxide. p + 502 — React tetraphosphorus decoxide with water to make phosphoric acid. + 61-120 4H3P04
  4. What is the minimum mass of water that must be added to 2.50 x 104 kg P 4010 to form phosphoric acid in the following reaction? p 4010 + 6H20 4H3P04
  5. Goal: To develop conversionfactors that will convert between a measurable propértye(mass} and number of Measurable Property 1 Number of Particles 1 Number of Particles 2 Measurable Property 2 Mass 1 Moles 1 Moles 2 Mass 2
  6. Two Very Similar Calculations • Conversion between amount of compound and amount of element in that compound (from Section 6.7 of the atoms- first version of the text and Section 9.4 of the chemistry- first version). 1 mol-P 283.889gP401() 1 kg (10 ? kg 104010 — 1.09 x 104 kg» I moVP4CYi7) 10B * 1 kg 30.97"! — 2.50 x 104 kgP401 • Conversion between units of one substance and units of another substance, both involved in a chemical reaction. 103 & 6 mol-H.0 18.0153gH20 1 kg ? kg - 2.50 x 1 kg 283.88981>4016 - 9.52 x 103 kg HA)
  7. Molar Ratio from Coefficients in Balanced Equations P4010 1 molecule P 4010 1 dozen P4010 molecules molecules P 4010 1 mole P4010 61-420 6 molecules H20 6 dozen 1-120 molecules 6(6.022x1023) molecules 1-120 6 moles 1420 H 3 p 04 4 molecules H3P04 4 dozen H3P04 molecules 4(6.022x1023) molecules H3P04 4 moles H3P04
  8. Calculations What is the minimum mass of water that must be added to 2.50 x 104 kg P 4010 to form phosphoric acid in the following reaction? p 4010 + 6H20 4H3P04 The coefficients in the balanced equation provide us with a conversion factor that converts from units of P4010 to units of 1-120. I mol P 4010 6 mol H20 I mol P401() 4 mol H3P04 6 mol H20 4 mol H3P04
  9. Sample Calculation What is the minimum mass of water that must be added to 2.50 x 104 kg P4010 to form phosphoric acid in the following reaction? p 4010 + 61-120 ? = 2.50 x 104 kg Jkg - 9.52 x kgH2 4H3P04 6 moVH20 18.0153 gH20 283.889gP40hiG 103 g
  10. Eq u ation Stoichiometry • Tip-off - The calculation calls for you to convert from amount of one substance to amount of another, both of which are involved in a chemical reaction. • General Steps 1. If you are not given it, write and balance the chemical equation for the reaction.
  11. Eq u ation Stoichiometry 2. Start your unit analysis in the usual way. 3. If you are no given grams of substance 1, convert from the unit that you are given to grams. This may take one or more conversion factors. 4. Convert from grams of substance 1 to moles of substance 1.
  12. Eq u ation Stoichiometry 5. Convert from moles of substance 1 to moles of substance 2 using the coefficients from the balanced equation to create the molar ratio used as a conversion factor. 6. Convert from moles of substance 2 to grams of substance 2, using its molar mass. 7. If necessary, convert from grams of 2 to the desired unit for 2. This may take one or more conversion factors.
  13. Equation Stoichiometry Steps Given unit Of substance I Unit analysis conversion factors Grams of substance I I mol I Using molar ass substahce 1 : (formula mass l) I Moles of substance (coeffic•ent 2) mol 2 Usin the mole ratio derived from the coe cients in the balanced equation: (coefficient l) mol I Moles of substance 2 (formula mass 2) g 2 Using molar mass Of substance 2: I mol 2 Grams Of substance 2 Unit analysis conversion factors Desired unit of substance 2
  14. Equation Stoichiometry Steps Molar mass of substance 2 (coefficient 2) mol 2 I mol I ? (unit) 2 = (given) (unit) I g V (coefficient l) moll I mol 2 (unit) (unit) One or more conversiopJactor Molar mass of convert the given unit t6gram substance I One or more conversion factors convert grams to the given unit.
  15. Sample Calculation Tetrachloroethene, C2C14, often called perchloroethylene (perc), is a colorless liquid used in dry cleaning. It can be formed in several steps from the reaction of dichloroethane, chlorine gas, and oxygen gas. What isye maximum mass of perchloroethylene, C2C14, that can be formed from 23.75 kilograms of dichloroethane, C2H4C12? The equation for the net reaction is: 8C2H4C12(/) + 6C12(g) + 702(g) -+ 4C2HC13(/) + 4C2C14(/) + 14H20(/)
  16. What is the maximum mass of perchloroethylene, C2C14, that can be formed from 23.75 kilograms of dichloroethane, C2H4C12? The equation for the net reaction is: 8C2H4C12(/) + 6C12(g) + 702(g) 4C2HC13(/) + 4C2C14(/) + 14H20(/) Usual unit analysis steps se ? kg C2C14 = 23.75Ag C2H4C12 kg IA«-xe Tip-off: The calculation calls for you to convert from amount of one substance to amount of another, both of which are involved in a chemical reaction.
  17. What is the maximum mass of perchloroethylene, C2C14, that can be formed from 23.75 kilograms of dichloroethane, C2H4C12? The equation for the net reaction is: 8C2H4C12(/) + 6C12(g) + 702(g) 4C2HC13(/) + 4C2C14(/) + 14H20(/) ? kg C2C14 = 23.75Ag c2V+4€h ? kg = 23.75 C2H4€v 2 ? kg C2C14 = 23.75Ag ? kg C2C14 = 23.75k C2Hq€b 103 g I kg 103 & I kg 1038 1038 1 mol C2H4C12 98.959#C2H4C12 98.959*C2H4C12 2 4 2)
  18. What is the maximum mass of perchloroethylene, C2C14, that can be formed from 23.75 kilograms of dichloroethane, C2H4C12? The equation for the net reaction is: 8C2H4C12(/) + 6C12(g) + 702(g) 4C2HC13(/) + 4C2C14(/) + 14H20(/) 103 & ? kg - 23.75Ag I kg 103 & kg - 23.754* ? kg 23.75AgC2H4Cti kg 103 & kg - 23.754* 103 & ? kg - 23.75Ag 4 mol 98.959vc2H4C12 8 98.959#C2H4C12 98.959#C2H4C12 98.959#C2H4C12 4 165.833 g I tnoLC2CEi4 I tnoLC2CT4 gC2C14 I kg 8 moVC#C13
  19. What is the maximum mass of perchloroethylene, C2C14, that can be formed from 23.75 kilograms of dichloroethane, C2H4C12? The equation for the net reaction is: 8C2H4C12(/) + 6C12(g) + 702(g) 4C2HC13(/) + 4C2C14(/) + 14H20(/) 165.833gC2C14 ? kg = 23.75Ag 19.90 kg 8 I mol-C2C17, 103 g
  20. Equation Stoichiometry Shortcut for Mass-Mass Problems Given mass of substance 1 (coefficient 2) (formula mass 2) (any mass unit) substance 2 Using (coefficient l) formula massA) (same mass unit) substance I Same mass unit of substance (coefficient 2) (formula mass 2) (any mass unit) substance 2 (unit) 2 — (given) (unit) I (coefficient l) (formula mass l) (same mass unit) substance I
  21. What is the maximum mass of perchloroethylene, C2C14, that can be formed from 23.75 kilograms of dichloroethane, C2H4C12? The equation for the net reaction is: 8C2H4C12(/) + 6C12(g) + 702(g) 4C2HC13(/) + 4C2C14(/) + 14H20(/) Usual unit analysis steps ? kg C2C14 23.75k C2H4C 2 kg Tip-off: The calculation calls for you to convert from mass of one substance to mass of another, both of which are involved in a chemical reaction. We can use the shortcut. 4x165.833 kg ? kg - 19.90 kg C2C14
  22. Questions to Ask. When Designing a Process for Making a Substance • How much of each reactant should be added to the reaction vessel? • What level of purity is desired for the final product? If the product is mixed with other substances (such as excess reactants), how will this purity be achieved?
  23. Limiting Component C) "Ihe wheels run out first, limiting the number of bicycles to six. The frames are in excess.
  24. Why limiting? (1) • To ensure that one reactant is converted to products most completely. — Expense P4(s) + 502(g) P4010(s) + excess 02(g) — Importance Si02(s) + 2C(s) Si(l) + 2CO(g) + excess C(s)
  25. Why limiting? (2) «•—..--m Concern for excess reactant that remains — danger pas) + 502(g) P4010(s) + excess 02(g) — ease of separation Si02(s) + 2C(s) Si(l) + 2CO(g) + excess C(s)
  26. Limiting Component C) "Ihe wheels run out first, limiting the number of bicycles to six. The frames are in excess.
  27. Component (2) ? bicycles — 7 frames ? bicycles = 12 wheels 1 bicycle I frame and 1 bicycle 2 wheels 1 bicycle = 7 bicycles 1 frame- 1 bicycle = 6 bicycles 2 wheels
  28. Reactant • The reactant that runs out first in a chemical reaction limits the amount of product that can form. This reactant is called the limiting reactant.
  29. Problems Tip-off - You are given amounts of two or more reactants in a chemical reaction, and you are asked to calculate the maximum amount of a product that can form from the combination of the reactants. General Steps 1. Do separate calculations of the maximum amount of product that can form from each reactant. 2. The lowest of the values calculated in the step above is your answer. It is the maximum amount of product that can be formed from the given amounts of reactants.
  30. Example The uranium(lV) oxide, IJ02, used as fuel in nuclear power plants has a higher percentage of the fissionable isotope uranium-235 than is present in the U02 found in nature. To make fuel grade U02, chemists first convert uranium oxides to uranium hexafluoride, UF6, whose concentration of uranium-235 can be increased by a process described in my online lecture for Section 16.3 of the atoms-first version of my text and Section 18.3 of the chemistry-first version. The enriched IJF6 is then converted back to I-J02 in a series of reactions, beginning with UF6 + 2H20 IJ02F2 + 41-IF a. How many megagrams of U02F2 can be formed from the reaction of 24.543 Mg IJF6 with 8.0 Mg of water? b. Why do you think the reactant in excess was chosen to be in excess?
  31. Example I-JF6 + 21-420 U02F2 + 4 HF How many megagrams of U02F2 can be formed from the reaction of 24.543 Mp IJF6 with 8.0 Mg of water? • Limiting reactant tip-off— We are given amounts of two reactants and asked to calculate an amount of product. Steps — Calculate amount of IJ02F2 from 24.543 Mg UF6. — Calculate amount of U02F2 from 8.0 Mg of 1-120. — Smaller value is answer.
  32. Example I-JF6 + 2H20 U02F2 + 41—IF How many megagrams of U02F2 can be formed from the reaction of 24.543 Mg IJF6 with 8.0 Mg of water? ? Mg UOÅF 24.543 UF6 ? Mg U02F2 = 24.543 Mg UF6 1 1 x 308.0245 ? Mg U02F2 = 24.543 Mg UF6 1 x 352.019
  33. Example I-JF6 + 2H20 U02F2 + 41—IF How many megagrams of U02F2 can be formed from the reaction of 24.543 Mg UF6 with 8.0 Mg of water? 1 x 308.0245 Mg ? Mg -24.543 Mg UF6 1 x 352.019 Mg 1 x 308.0245 Mg U02F2 ? Mg U02F2 - - 24.543 1 x 352.019 Mg-em 1 x 308.0245 Mg U02F2 = 24.543 Mg-em ? Mg U02F2 1 x 352.019 - 21.476 Mg U02F2
  34. Example UF6 + 2H20 ? Mg U02F2 = 8.0 IJ02F2 + 4 HF How many megagrams of IJ02F2 can be formed from the reaction of 24.543 Mg UF6 with 8.0 Mg of water? 1 x 308.0245 Mg U02F2 ? Mg U02F 'L 24.543 1 x 352.019 Mg-em = 21.476 Mg U02F2 1 x 308.0245 Mg U02F2 2 x 18.0153 = 68 Mg U02F2
  35. Example b. Why do you think the reactant in excess was chosen to be in excess? — Water is much less expensive than the rare uranium compound. — It is important to carry as much uranium through the process as possible. — It is better to have the nontoxic water in the final mixture than the radioactive UF6. — Water is also easy to separate from the solid product mixture.
  36. not 1000/0 Yield? Reversible reactions Side reactions Slow reactions Loss during separation/purification
  37. Percent Yield Actual Yield Percent Yield = x 100% Thyoretical Yield Actual yield is measured. It is given in the problem. Theoretical yield is the maximum yield that you calculate.
  38. Example Calculation for Percent Yield Sodium chromate, Na2Cr04, is made by roasting chromite, FeCr204, with sodium carbonate, Na2C03. (Roasting means heating in the presence of air or oxygen.) A simplified version of the net reaction is 4FeCr204 + 8Na2C03 + 702 8Na2Cr04 + 2Fe203 + 8C02 What is the percent yield if 1.2 kg of Na2Cr04 is produced from ore that contains 1.0 kg of FeCr204?
  39. Example Calculation for Percent Yield 4FeCr204 + 8Na2C03 + 702 8Na2Cr04 + 2Fe203 + 8C02 What is the percent yield if 1.2 kg of Na2Cr04 is produced from ore that contains 1.0 kg of FeCr204? Actual Yield Percent Yield = x 1000/0 Theoretical Yield — Na2Cr04 is a product, so 1.2 kg Na2Cr04 is the actual yield. — FeCr204 is a reactant, so 1.0 kg FeCr204 is used to calculate the maximum amount of Na2Cr04 that could be formed (the theoretical yield).
  40. Example Calculation for Percent Yield 4FeCr204 + 8Na2C03 + 702 8Na2Cr04 + 2Fe203 + 8C02 What is the percent yield if 1.2 kg of Na2Cr04 is produced from ore that contains 1.0 kg of FeCr204? 1 8 moVNa2CrOQ 161.9733 Na2Cr04 103 g ? kg Na2C104 1.0-kg-FeCtä04 Amol I moLNa2Cr04 223.835* 1.4 kg Na2Cr04 Actual yield Percent yield = x 100% "Iheoretical yield 1.2 kg Na2Cr04 x 100% - 1.4 kg Na2Cr04 1 kg 103 g — 86% yield
  41. to Moles Measurable property Mass Volume of solution Moles Moles of pure substance Moles of solute
  42. Molarity moles of solute Molarity liter of solution Converts between moles of solute and volume of solution 0.500 mol Na3P04 0.500 mol Na3P04 0.500 M Na3P04 means or 103 mL Na3P04 solution 1 L Na3P04 solution
  43. Calculating Molarity • A silver perchlorate solution was made by dissolving 29.993 g of pure AgC104 in water and then diluting the mixture with additional water to achieve a total volume of 50.00 mL. What is the solution's molarity?
  44. Calculating Molarity A silver perchlorate solution was made by dissolving 29.993 g of pure AgC104 in water and then diluting the mixture with additional water to achieve a total volume of 50.00 mL. What is the solution's molarity? ? mol AgC104 ? M AgC104 I L AgC104 soln ? mol AgClÖ 29.993 g AgC104 ? M AgC104 I L AgClQ4 soln ? mol AgC104 ? M AgC104 I L AgC104 soln ? mol AgC104 ? M AgC104 I L AgC104 soln 50.00 mL AgC104 soln 29.993g-AgGIOw 50.00 mL AgC104 soln 29.993&AgClOa- 50.00 mL AgC104 soln 4 I mol AgC104 207.3185 &AgGIO-c
  45. Calculating Molarity A silver perchlorate solution was made by dissolving 29.993 g of pure AgC104 in water and then diluting the mixture with additional water to achieve a total volume of 50.00 mL. What is the solution's molarity? ? M AgC104 - ? M AgC104 ? M AgC104 ? mol AgC104 I L AgC104 soln ? mol Age 4 I L AgC104 soln ? mol AgC104 I L AgC,lO/å soln 2.893 mol AgC104 1 L AgC104 soln 29.993gAgCIOe ml;AgClOvsoln 29.993gAgCIC)v 30.00 ml; AgC104 soln 29.993&AgG1€v 50.00 mEAgC104 soln - 2.893 M AgC104 CiQ mol AgClO/1 207.3185 gAgCIO-c I mol AgC104 207.3185 g-AgClO-„- I mol AgC104 207.3185KAgGIOö 103 IL 103 ml: IL
  46. Calculating Molarity A silver perchlorate solution was made by dissolving 29.993 g of pure AgC104 in water and then diluting the mixture with additional water to achieve a total volume of 50.00 mL. What is the solution's molarity? Molarity expressed with more specific uh ? mol AgC104 ? M AgC104 iven amount of solute 29.993gAgGIOw Converts the given Converts mass volume unit into the to moles desired volume unit 1 mol AgC104 103 mE 1 LAgC104 soln 50.00 mEAgC104 soln 207.3185gAgC}Ov 1 L Given amount of solution 2.893 mol AgC104 2.893 M AgC104 1 L AgC104 soln
  47. Equation Stoichiometry Start here when mass of pure substance is given. -g any mass unit 1 (any mass unit) grams I I mol I (formula mass) g I moles I (number from molariry) mol h 1 L (or 103 mL) soln I (L or ml,) Of solution I (L or mL) (any volume unit) any volume unit Of solution I any mass unit 2 (any mass unit) grams 2 (formula mass) g 2 I mol 2 Can be converted (coef. ) r(soef. V) Elliis is the core of any equation stoichiometry problem. moles 2 into mass or into volume of solution I L (or 103 mL) soln 2 (number from molarity) mol 2 (L or mL) Of solution 2 (any volume unit) (L or mL) any volume unit Of solution 2 Start here when volume of solution is given.
  48. Eq u ation Stoichiometry • Tip-off - The calculation calls for you to convert from amount of one substance to amount of another, both of which are involved in a chemical reaction. • General Steps 1. If you are not given it, write and balance the chemical equation for the reaction. 2. Start your unit analysis in the usual way.
  49. Eq u ation Stoichiometry 3. Convert from the units that you are given for substance 1 to moles of substance 1. — For pure solids and liquids, this means converting mass to moles using the molar mass of the substance. — Molarity can be used to convert from volume of solution to moles of solute.
  50. Equation Stoichiometry 4. 5. Convert from moles of substance 1 to moles of substance 2, using the coefficients in the balanced equation. Convert from moles of substance 2 to the desired units for substance 2. For pure solids and liquids, this means converting moles to mass using the molar mass of substance 2. Molarity can be used to convert from moles of solute to volume of solution.
  51. Equation Stoichiometry 6. If necessary, we convert from grams to the mass unit we want, or liters (or milliliters) to the volume unit we want. 7. Calculate your answer and report it with the correct significant figures (in scientific notation, if necessary) and unit.
  52. Example 1 • How many milliliters of 6.00 M HN03 are necessary to neutralize the carbonate in 75.00 mL of 0.250 M Na2C03? 0.250 mol Na2C03 103 mL Na2C03 soln 103 mL HN03 soln 6.00 mol HN03
  53. Example 1 How many milliliters of 6.00 M HN03 are necessary to neutralize the carbonate in 75.00 mL of 0.250 M Na2C03? • Before we can do our calculation, we need to get a balanced equation. HN03 is a strong acid and Na2C03 as a weak base, so we follow the steps described in a previous lesson for writing double displacement equations for acid-base reactions. 2HN03(aq) + Na2C03(aq) H20(l) + C02(g) + 2NaN03(aq)
  54. Example 1 • How many milliliters of 6.00 M HN03 are necessary to neutralize the carbonate in 75.00 mL of 0.250 M Na2C03? 2HN03(aq) + Na2C03(aq) -9 H20(l) + C02(g) + 2NaN03(aq) ? mL HN03 soln ? HN03 soln ? mL HN03 soln ? mL HN03 soln - 75.00 3 0.250 mol Na2C03 75.00 0.250 75.00 0.230 moVNa2C03 = 75.00 ml-Na2COss01n 103 mL&2COssohi 2 mol HN03 I -mol-Na2€03
  55. Example 1 • How many milliliters of 6.00 M HN03 are necessary to neutralize the carbonate in 75.00 mL of 0.250 M Na2C03? 0.250 mobNa2€Og ? ml. HN03 soln - 75.00 10-3 mLaNa2COesohi 0.250 moVNa2COg ? ml. H NO; soln — - 75.00 1 03 ml—N7VCOysoht 0.250 ,cog ? HN03 SOIn - 75.00 103 ml—Xia2COs-soin 6.25 HN03 soln 2 mol+fNC)j- 2 mol+1NC-Y€ 1 -mol-Na2CCY3 103 ml. HNOS soln 6.00 m0144NOs 103 ml. HNOS soln 6.00-mol-HN03 Coefficients from balanced equation convert moles of one substance into moles of another substance. 0.250 2 mol+1NOö 103 ml. HNOS soln ? mL HN03 soln — 75.00 14nolAa2C03 6.00 6.25 mL HNOS soln Molarity as a conversion MO arity as a conversion factor converts liters into moles. factor converts moles into liters.
  56. Example 2 What is the maximum number of grams of silver chloride that will precipitate from a solution made by mixing 25.00 mL of 0.050 M MgC12 with an excess of AgN03 solution? 0.050 mol MgC12 103 mL MgC12 soln mol AgN03 gAgN03
  57. Example 2 What is the maximum number of grams of silver chloride that will precipitate from a solution made by mixing 25.00 mL of 0.050 M MgC12 with an excess of AgN03 solution? • Before we can do our calculation, we need to get a balanced equation. This is a precipitation reaction, so we follow the steps described in a previous lesson for writing double displacement equations for precipitation reactions. 2AgN03(aq) + MgC12(aq) 2AgCl(s) +
  58. Example 2 What is the maximum number of grams of silver chloride that will precipitate from a solution made by mixing 25.00 mL of 0.050 M MgC12 with an excess of AgN03 solution? 2AgN03(aq) + MgC12(aq) 2AgCl(s) + g AgCl g Agc = 25.00 mL-MgGEscjlh 2 0.050 mol MgC12 = 25.00 103 mL-MgGlj-sOY 0.050 g AgC1 = 25.00 103 m-L-MgGb-sotn 0.050 g AgC1 = 25.00 103 mL-MgGly-SOif1 2 mol AgCl
  59. Example 2 What is the maximum number of grams of silver chloride that will precipitate from a solution made by mixing 25.00 mL of 0.050 M MgC12 with an excess of AgN03 solution? 0.050 g AgCl = 25.00 ml—Mg€lYßOlii 103 0.050 g AgCl = 25.00 ml—Mg€4ysOlii 103 ml—Mg€ffSSOtfi 0.050 moVMg€_ibi g AgC1 25.00 103 ml—Mg€ffSSOlt1 0.36 g AgC1 mOV-Ag€iF mol-Mg€;h- 1 mol-Mg€ffs 2 molAger molaAg€t 143.3209 g AgCl 1 mol-fig€t 143.3209 g AgC1 I mol-Ag€„t

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