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Stoichiometry 3. Gas and Solution Stoichiometry / Titration. Review. So far we’ve learned that you can convert from moles of one substance to moles of any other substance in one step. Provided you have a balanced chemical equation.
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Stoichiometry 3 Gas and Solution Stoichiometry / Titration
Review • So far we’ve learned that you can convert from moles of one substance to moles of any other substance in one step. • Provided you have a balanced chemical equation. • You can also start stoichiometry problems with grams and particles (atoms, molecules, ions, etc) provided you have the appropriate conversion factors. • 1 mol = 6.022x1023 particles • 1 mol = (molar mass) grams • The molar mass is unique to every substance. • Have to work it out on a case by case basis.
particles A particles A g A g B mol A mol B L A(g) L B(g) L A(aq) L B(aq) Stoichiometry Map 1 mol = 6.022x1023 particles 1 mol = 6.022x1023 particles 1 mol = (molar mass) g 1 mol = (molar mass) g BCE 1 mol = 22.4 L 1 mol = 22.4 L (Molarity) mol = 1 L (Molarity) mol = 1 L
Gas Stoichiometry • How many liters of CO2 (at STP) are produced by the decomposition of 14.0 g CaCO3 in the following reaction? • CaCO3(s) CaO(s) + CO2(g) • Given: 14.0 g CaCO3 • Want: ??? L CO2 • Conversion factors: • 1 mol CaCO3 = 100.09 g CaCO3 • 1 mol CaCO3 yields 1 mol CO2 • At STP, 1 mol CO2 = 22.4 L CO2 • Plan: 14.0 g CaCO3 mol CaCO3 mol CO2 L CO2 14.0 g CaCO3 = 3.13 L CO2
Gas Stoichiometry • If 25.0 mL of H2 (at STP) are available to react with excess Fe2O3, how many grams of Fe can be produced? • Fe2O3(s) + 3H2(g) 2Fe(s) + 3H2O(l) • Given: 25.0 mL H2 • Want: ??? g Fe • Conversion factors: • 0.001 L = 1 mL • At STP, 1 mol H2 = 22.4 L H2 • 3 mol H2 yield 2 mol Fe • 1 mol Fe = 55.85 g Fe • Plan: 25.0 mL H2 L H2 mol H2 mol Fe g Fe 25.0 mL H2 = 0.0416 g Fe
Gas Stoichiometry • How to simplify volume-volume problems. • If the given and wanted values are both in liters (or both in milliliters) you can simplify the problem a bit. • Why? • Because you’d have to divide by 22.4 to get from liters to moles, then multiply by 22.4 again to get from moles to liters on the other side. • If you start and end with milliliters, the same thing happens with the mL L conversion on one side and the L mL conversion on the other side. • Since those operations cancel each other out, you can bypass them if you’re cautious!
Gas Stoichiometry • What volume of hydrogen gas (at STP) is needed to react with 20.0 L of N2 gas (at STP) in the following reaction? • N2(g) + 3H2(g) 2NH3(g) • Given: 20.0 L N2 • Want: ??? L H2 • Conversion factors: • 1 mol N2 reacts with 3 mol H2 • At STP, 1 mol N2 = 22.4 L N2 • At STP, 1 mol H2 = 22.4 L H2 • Since both moles translate to liters in the same way, we can bypass moles and convert from L N2 L H2 directly. 20.0 L N2 = 60.0 L H2
Solution Stoichiometry • In solution stoichiometry we use the molar concentrations of solutions as conversion factors from liters to moles. • (Molarity) moles = 1 liter • In a 2.50 M HCl solution, there are 2.50 mol HCl dissolved in 1 L of solution.
Solution Stoichiometry • How many grams of aluminum are needed to react with 200. mL of 0.250 M CuCl2? • 2Al(s) + 3CuCl2(aq) 2AlCl3(aq) + 3Cu(s) • Given: 200. mL CuCl2 • Want: ??? g Al • Conversion factors: • 0.001 L = 1 mL • 0.250 mol CuCl2 = 1 L • 2 mol Al reacts with 3 mol CuCl2 • 1 mol Al = 26.98 g Al • Plan: 200. mL CuCl2 L CuCl2 mol CuCl2 mol Al g Al 200. mL CuCl2 = 0.899 g Al
Solution Stoichiometry • What volume of 1.25 M NaOH is required to react with 0.800 L of 1.90 M H2SO4? • 2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l) • Given: 0.800 L H2SO4 • Want: ??? L NaOH • Conversion factors: • 1.90 mol H2SO4 = 1 L H2SO4 • 2 mol NaOH reacts with 1 mol H2SO4 • 1.25 mol NaOH = 1 L NaOH • Plan: 0.800 L H2SO4 mol H2SO4 mol NaOH L NaOH 0.800 L H2SO4 = 2.43 L NaOH
Titrations • A titration is a quantitative experiment designed to determine the unknown concentration of a solution by reacting it with another solution of known concentration. • Titrant – the solution with known concentration. • Analyte – the solution with unknown concentration. • Chemists add the titrant solution to the analyte solution until an equivalence point is reached. • Equivalence point – the point where the amount of titrant added is stoichiometrically equal to the amount of analyte present. • Depending on the reaction, there may be more than one equivalence point. • Chemists use indicators or meters to determine when the equivalence point has been reached.
Titrations • 24.5 mL of an HCl solution with unknown concentration is titrated with 30.2 mL of 0.105 M NaOH solution. What is the molar concentration of the HCl solution? • HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) • Molarity = mol solute/L solution • We already know the volume of the solution: 24.5 mL • We can convert 24.5 mL L • Molarity = mol solute / 0.0245 L solution 24.5 mL HCl = 0.0245 L HCl
Titrations • Now we have to determine how many moles of HCl were in the original analyte sample. • Given: 30.2 mL of NaOH solution used • Want: ??? mol HCl in original sample • Conversion factors: • 0.001 L NaOH = 1 L NaOH • 0.105 mol NaOH = 1 L NaOH • 1 mol NaOH reacts with 1 mol HCl • Plan: 30.2 mL NaOH L NaOH mol NaOH mol HCl 30.2 mL NaOH = 0.00317 mol HCl
Titrations • The molar concentration of the HCl solution is: • M = 0.00317 mol HCl / 0.0245 L solution = 0.129 M