Comprehensive Chemical Stoichiometry, Yield Analysis, and Volumetric Titrations

Fundamental Mole Calculations and Chemical Composition

  • Molar Mass of Dimethylamine, (CH3)2NH(CH_3)_2NH:

    • Dimethylamine is a secondary amine composed of two methyl groups attached to a nitrogen atom.
    • Summing atomic masses yields:
      • Carbon: 2×12.011 g mol−1=24.022 g mol−12 \times 12.011\,g\,mol^{-1} = 24.022\,g\,mol^{-1}
      • Hydrogen: 7×1.008 g mol−1=7.056 g mol−17 \times 1.008\,g\,mol^{-1} = 7.056\,g\,mol^{-1}
      • Nitrogen: 1×14.007 g mol−1=14.007 g mol−11 \times 14.007\,g\,mol^{-1} = 14.007\,g\,mol^{-1}
    • Total molar mass of (CH3)2NH=45.0 g mol−1(CH_3)_2NH = 45.0\,g\,mol^{-1}.

  • Mole Determination in Iron(II) Sulfate, FeSO4FeSO_4:

    • Given mass of FeSO4=143.0 gFeSO_4 = 143.0\,g.
    • Molar mass of FeSO4=55.845 g mol−1+32.065 g mol−1+4(15.999 g mol−1)=151.91 g mol−1FeSO_4 = 55.845\,g\,mol^{-1} + 32.065\,g\,mol^{-1} + 4(15.999\,g\,mol^{-1}) = 151.91\,g\,mol^{-1}.
    • Calculating moles via dimensional analysis:
      • Moles of FeSO4=143.0 g151.91 g mol−1=0.9414 mol FeSO4\text{Moles of } FeSO_4 = \frac{143.0\,g}{151.91\,g\,mol^{-1}} = 0.9414\,mol\,FeSO_4

  • Carbon Atom Enumeration in Morphine, C17H19NO3C_{17}H_{19}NO_3:

    • Molecular formula: C17H19NO3C_{17}H_{19}NO_3
    • Sample mass: 53.8 g53.8\,g
    • Molar mass calculation:
      • Molar mass=17(12.011)+19(1.008)+14.007+3(15.999)=285.34 g mol−1\text{Molar mass} = 17(12.011) + 19(1.008) + 14.007 + 3(15.999) = 285.34\,g\,mol^{-1}
    • Conversion sequence from mass to molecules to carbon atoms:
      • Moles of morphine=53.8 g285.34 g mol−1=0.1885 mol\text{Moles of morphine} = \frac{53.8\,g}{285.34\,g\,mol^{-1}} = 0.1885\,mol
      • Molecules of morphine=0.1885 mol×(6.022×1023 molecules mol−1)=1.135×1023 molecules\text{Molecules of morphine} = 0.1885\,mol \times (6.022 \times 10^{23}\,\text{molecules}\,mol^{-1}) = 1.135 \times 10^{23}\,\text{molecules}
      • Number of C atoms=1.135×1023 molecules×17 C atoms/molecule=1.93×1024 C atoms\text{Number of C atoms} = 1.135 \times 10^{23}\,\text{molecules} \times 17\,\text{C atoms/molecule} = 1.93 \times 10^{24}\,\text{C atoms}

  • Mass of Avogadro's Number of Raindrops:

    • Average raindrop mass: 0.063 g0.063\,g
    • Avogadro's Number: 6.022×1023 raindrops6.022 \times 10^{23}\,\text{raindrops}
    • Unit conversion factor: 1 metric ton=1000 kg=1.0×106 g1\,\text{metric ton} = 1000\,kg = 1.0 \times 10^6\,g
    • Calculated mass:
      • Total mass=3.8 metric tons\text{Total mass} = 3.8\,\text{metric tons}

  • Mass Calculation of Dinitrogen Pentoxide, N2O5N_2O_5:

    • Given amount: 1.75 mol N2O51.75\,mol\,N_2O_5
    • Molar mass of N2O5=2(14.007 g mol−1)+5(15.999 g mol−1)=108.01 g mol−1N_2O_5 = 2(14.007\,g\,mol^{-1}) + 5(15.999\,g\,mol^{-1}) = 108.01\,g\,mol^{-1}
    • Mass calculation:
      • Mass=1.75 mol×108.01 g mol−1=189 g N2O5\text{Mass} = 1.75\,mol \times 108.01\,g\,mol^{-1} = 189\,g\,N_2O_5

  • Percentage Composition of Streptomycin, C21H29O12N7C_{21}H_{29}O_{12}N_7:

    • Molar mass calculation:
      • Carbon: 21×12.011 g mol−1=252.23 g mol−1\text{Carbon: } 21 \times 12.011\,g\,mol^{-1} = 252.23\,g\,mol^{-1}
      • Hydrogen: 29×1.008 g mol−1=29.23 g mol−1\text{Hydrogen: } 29 \times 1.008\,g\,mol^{-1} = 29.23\,g\,mol^{-1}
      • Oxygen: 12×15.999 g mol−1=191.99 g mol−1\text{Oxygen: } 12 \times 15.999\,g\,mol^{-1} = 191.99\,g\,mol^{-1}
      • Nitrogen: 7×14.007 g mol−1=98.05 g mol−1\text{Nitrogen: } 7 \times 14.007\,g\,mol^{-1} = 98.05\,g\,mol^{-1}
      • Total molar mass=571.50 g mol−1\text{Total molar mass} = 571.50\,g\,mol^{-1}
    • Mass percentages:
      • %C=252.23571.50×100%=44.13%\%C = \frac{252.23}{571.50} \times 100\% = 44.13\%
      • %H=29.23571.50×100%=5.08%\%H = \frac{29.23}{571.50} \times 100\% = 5.08\%
      • %O=191.99571.50×100%=33.63%\%O = \frac{191.99}{571.50} \times 100\% = 33.63\%
      • %N=98.05571.50×100%=17.2%\%N = \frac{98.05}{571.50} \times 100\% = 17.2\%

  • Empirical and Molecular Formulas of Nicotine:

    • Description: Nicotine is a poisonous compound found in tobacco leaves.
    • Elemental mass percentages: 74.0% C74.0\%\,C, 8.65% H8.65\%\,H, and 22.3% N22.3\%\,N
    • Molar mass: 162 g mol−1162\,g\,mol^{-1}
    • Formula determinations recorded:
      • EF=CHN\text{EF} = CHN
      • EF=CHN\text{EF} = CHN
    • Analytical stoichiometric evaluation based on 100 g100\,g sample:
      • Moles of C=74.0 g12.011 g mol−1=6.16 mol\text{Moles of C} = \frac{74.0\,g}{12.011\,g\,mol^{-1}} = 6.16\,mol
      • Moles of H=8.65 g1.008 g mol−1=8.58 mol\text{Moles of H} = \frac{8.65\,g}{1.008\,g\,mol^{-1}} = 8.58\,mol
      • Moles of N=22.3 g14.007 g mol−1=1.59 mol\text{Moles of N} = \frac{22.3\,g}{14.007\,g\,mol^{-1}} = 1.59\,mol
      • Dividing by the lowest mole value (1.591.59) gives relative ratios yielding the empirical formula C5H7NC_5H_7N (empirical mass=81.13 g mol−1\text{empirical mass} = 81.13\,g\,mol^{-1}) and molecular formula factor 162 g mol−181.13 g mol−1≈2\frac{162\,g\,mol^{-1}}{81.13\,g\,mol^{-1}} \approx 2, corresponding to C10H14N2C_{10}H_{14}N_2.

  • Empirical and Molecular Formulas of Cumene:

    • Description: Cumene is a hydrocarbon composed exclusively of carbon and hydrogen.
    • Elemental composition: 89.94% carbon89.94\%\,\text{carbon}, which requires 10.06% hydrogen10.06\%\,\text{hydrogen}.
    • Molar mass: 120.2 g mol−1120.2\,g\,mol^{-1}
    • Stoichiometric derivation based on 100 g100\,g sample:
      • Moles of C=89.94 g12.011 g mol−1=7.488 mol\text{Moles of C} = \frac{89.94\,g}{12.011\,g\,mol^{-1}} = 7.488\,mol
      • Moles of H=10.06 g1.008 g mol−1=9.980 mol\text{Moles of H} = \frac{10.06\,g}{1.008\,g\,mol^{-1}} = 9.980\,mol
      • Mole ratio: 9.9807.488=1.333=43\text{Mole ratio: } \frac{9.980}{7.488} = 1.333 = \frac{4}{3}
    • Empirical Formula (EF\text{EF}):
      • EF=C3H4\text{EF} = C_3H_4
      • Empirical formula mass=3(12.011)+4(1.008)=40.065 g mol−1\text{Empirical formula mass} = 3(12.011) + 4(1.008) = 40.065\,g\,mol^{-1}
    • Molecular Formula (MF\text{MF}):
      • Scaling factor n=120.2 g mol−140.065 g mol−1=3\text{Scaling factor } n = \frac{120.2\,g\,mol^{-1}}{40.065\,g\,mol^{-1}} = 3
      • MF=C9H12\text{MF} = C_9H_{12}

Stoichiometric Relationships in Chemical Reactions

Stoichiometric mole conversions for chemical reactions

  • Stoichiometric Mole-to-Mole Equivalences:

    • Reaction (a) — Carbon Disulfide Synthesis:

      • Balanced chemical equation:
        • 4 C+S8→4 CS24\,C + S_8 \rightarrow 4\,CS_2
      • Problem statement: Determine the value of xx (in molmol) when 1.5 mol S81.5\,mol\,S_8 reacts to yield x mol CS2x\,mol\,CS_2.
      • Stoichiometric ratio:
        • 4 mol CS21 mol S8\frac{4\,mol\,CS_2}{1\,mol\,S_8}
      • Calculation:
        • x=1.5 mol S8×4 mol CS21 mol S8=6.0 mol CS2x = 1.5\,mol\,S_8 \times \frac{4\,mol\,CS_2}{1\,mol\,S_8} = 6.0\,mol\,CS_2

    • Reaction (b) — Carbon Disulfide Combustion:

      • Balanced chemical equation:
        • CS2+3 O2→CO2+2 SO2CS_2 + 3\,O_2 \rightarrow CO_2 + 2\,SO_2
      • Problem statement: Determine the value of xx (in molmol) when 3.7 mol CS23.7\,mol\,CS_2 reacts to yield x mol SO2x\,mol\,SO_2.
      • Stoichiometric ratio:
        • 2 mol SO21 mol CS2\frac{2\,mol\,SO_2}{1\,mol\,CS_2}
      • Calculation:
        • x=3.7 mol CS2×2 mol SO21 mol CS2=7.4 mol SO2x = 3.7\,mol\,CS_2 \times \frac{2\,mol\,SO_2}{1\,mol\,CS_2} = 7.4\,mol\,SO_2

    • Reaction (c) — Hydrazine Oxidation:

      • Balanced chemical equation:
        • N2H4+3 O2→2 NO2+2 H2ON_2H_4 + 3\,O_2 \rightarrow 2\,NO_2 + 2\,H_2O
      • Problem statement: Determine the value of xx (in molmol) when 3.1 mol O23.1\,mol\,O_2 reacts to yield x mol NO2x\,mol\,NO_2.
      • Stoichiometric ratio:
        • 2 mol NO23 mol O2\frac{2\,mol\,NO_2}{3\,mol\,O_2}
      • Calculation:
        • x=3.1 mol O2×2 mol NO23 mol O2=2.067 mol≈2.1 mol NO2x = 3.1\,mol\,O_2 \times \frac{2\,mol\,NO_2}{3\,mol\,O_2} = 2.067\,mol \approx 2.1\,mol\,NO_2

    • Reaction (d) — Silane Oxidation:

      • Balanced chemical equation:
        • SiH4+2 O2→SiO2+2 H2OSiH_4 + 2\,O_2 \rightarrow SiO_2 + 2\,H_2O
      • Problem statement: Determine the value of xx (in molmol) when 3.9×10−3 mol O23.9 \times 10^{-3}\,mol\,O_2 reacts to yield x mol H2Ox\,mol\,H_2O.
      • Stoichiometric ratio:
        • 2 mol H2O2 mol O2=1\frac{2\,mol\,H_2O}{2\,mol\,O_2} = 1
      • Calculation:
        • x=3.9×10−3 mol O2×2 mol H2O2 mol O2=3.9×10−3 mol H2Ox = 3.9 \times 10^{-3}\,mol\,O_2 \times \frac{2\,mol\,H_2O}{2\,mol\,O_2} = 3.9 \times 10^{-3}\,mol\,H_2O
        • Numerical output format: 3.9e−3 mol3.9e-3\,mol

Limiting Reactants and Theoretical Yields

Limiting reactant and theoretical yield problems

  • Reduction of Iron Ore to Metallic Iron:

    • Chemical process: Conversion of hematite ore into iron metal using carbon monoxide as a reducing agent.
    • Balanced equation:
      • Fe2O3(s)+3 CO(g)→2 Fe(s)+3 CO2(g)Fe_2O_3(s) + 3\,CO(g) \rightarrow 2\,Fe(s) + 3\,CO_2(g)
    • Initial conditions: 7.25 kg7.25\,kg of Fe2O3Fe_2O_3 and 7.25 kg7.25\,kg of COCO.
    • Determination of limiting reactant:
      • Molar mass of Fe2O3=2(55.845)+3(15.999)=159.69 g mol−1\text{Molar mass of } Fe_2O_3 = 2(55.845) + 3(15.999) = 159.69\,g\,mol^{-1}
      • Molar mass of CO=12.011+15.999=28.01 g mol−1\text{Molar mass of } CO = 12.011 + 15.999 = 28.01\,g\,mol^{-1}
      • Moles of Fe2O3=7250 g159.69 g mol−1=45.40 mol\text{Moles of } Fe_2O_3 = \frac{7250\,g}{159.69\,g\,mol^{-1}} = 45.40\,mol
      • Moles of CO=7250 g28.01 g mol−1=258.84 mol\text{Moles of } CO = \frac{7250\,g}{28.01\,g\,mol^{-1}} = 258.84\,mol
      • Stoichiometric requirement: 45.40 mol Fe2O345.40\,mol\,Fe_2O_3 requires 3×45.40=136.2 mol CO3 \times 45.40 = 136.2\,mol\,CO.
      • Because 258.84 mol CO258.84\,mol\,CO is present, COCO is in excess and Fe2O3Fe_2O_3 is the limiting reactant.
    • Maximum theoretical mass of iron (FeFe) produced:
      • Theoretical moles of Fe=45.40 mol Fe2O3×2 mol Fe1 mol Fe2O3=90.80 mol Fe\text{Theoretical moles of } Fe = 45.40\,mol\,Fe_2O_3 \times \frac{2\,mol\,Fe}{1\,mol\,Fe_2O_3} = 90.80\,mol\,Fe
      • Mass of Fe=90.80 mol×55.845 g mol−1=5070.7 g=5.07 kg Fe\text{Mass of } Fe = 90.80\,mol \times 55.845\,g\,mol^{-1} = 5070.7\,g = 5.07\,kg\,Fe

  • Oxidation of Ammonia in Nitric Acid Manufacture:

    • Chemical context: Catalytic oxidation step in the industrial manufacture of nitric acid.
    • Balanced equation:
      • 4 NH3(g)+5 O2(g)→4 NO(g)+6 H2O(g)4\,NH_3(g) + 5\,O_2(g) \rightarrow 4\,NO(g) + 6\,H_2O(g)
    • Initial conditions: 54.5 kg NH354.5\,kg\,NH_3 reacts with 118 kg O2118\,kg\,O_2.
    • Determination of limiting reactant:
      • Molar mass of NH3=14.007+3(1.008)=17.031 g mol−1\text{Molar mass of } NH_3 = 14.007 + 3(1.008) = 17.031\,g\,mol^{-1}
      • Molar mass of O2=2(15.999)=31.998 g mol−1\text{Molar mass of } O_2 = 2(15.999) = 31.998\,g\,mol^{-1}
      • Moles of NH3=54500 g17.031 g mol−1=3200.0 mol\text{Moles of } NH_3 = \frac{54500\,g}{17.031\,g\,mol^{-1}} = 3200.0\,mol
      • Moles of O2=118000 g31.998 g mol−1=3687.7 mol\text{Moles of } O_2 = \frac{118000\,g}{31.998\,g\,mol^{-1}} = 3687.7\,mol
      • Reaction equivalents:
        • For NH3:3200.04=800.0 equivalents\text{For } NH_3: \frac{3200.0}{4} = 800.0\,\text{equivalents}
        • For O2:3687.75=737.54 equivalents\text{For } O_2: \frac{3687.7}{5} = 737.54\,\text{equivalents}
      • Because 737.54<800.0737.54 < 800.0, oxygen (O2O_2) is the limiting reactant.
    • Theoretical yield of nitric oxide (NONO):
      • Moles of NO=3687.7 mol O2×4 mol NO5 mol O2=2950.16 mol NO\text{Moles of } NO = 3687.7\,mol\,O_2 \times \frac{4\,mol\,NO}{5\,mol\,O_2} = 2950.16\,mol\,NO
      • Molar mass of NO=14.007+15.999=30.006 g mol−1\text{Molar mass of } NO = 14.007 + 15.999 = 30.006\,g\,mol^{-1}
      • Mass of NO=2950.16 mol×30.006 g mol−1=88523 g=88.5 kg NO\text{Mass of } NO = 2950.16\,mol \times 30.006\,g\,mol^{-1} = 88523\,g = 88.5\,kg\,NO

  • Oxidation of Copper Metal by Nitric Acid:

    • Balanced equation:
      • 3 Cu(s)+8 HNO3(aq)→3 Cu(NO3)2(aq)+2 NO(g)+4 H2O(l)3\,Cu(s) + 8\,HNO_3(aq) \rightarrow 3\,Cu(NO_3)_2(aq) + 2\,NO(g) + 4\,H_2O(l)
    • Initial conditions: 24.3 g Cu24.3\,g\,Cu reacts with 59.6 g HNO359.6\,g\,HNO_3.
    • Determination of limiting reactant:
      • Molar mass of Cu=63.546 g mol−1\text{Molar mass of } Cu = 63.546\,g\,mol^{-1}
      • Molar mass of HNO3=1.008+14.007+3(15.999)=63.012 g mol−1\text{Molar mass of } HNO_3 = 1.008 + 14.007 + 3(15.999) = 63.012\,g\,mol^{-1}
      • Moles of Cu=24.3 g63.546 g mol−1=0.3824 mol\text{Moles of } Cu = \frac{24.3\,g}{63.546\,g\,mol^{-1}} = 0.3824\,mol
      • Moles of HNO3=59.6 g63.012 g mol−1=0.9459 mol\text{Moles of } HNO_3 = \frac{59.6\,g}{63.012\,g\,mol^{-1}} = 0.9459\,mol
      • Reaction equivalents:
        • For Cu:0.38243=0.1275 equivalents\text{For } Cu: \frac{0.3824}{3} = 0.1275\,\text{equivalents}
        • For HNO3:0.94598=0.1182 equivalents\text{For } HNO_3: \frac{0.9459}{8} = 0.1182\,\text{equivalents}
      • Because 0.1182<0.12750.1182 < 0.1275, nitric acid (HNO3HNO_3) is the limiting reactant.
    • Mass of nitric oxide (NONO) formed:
      • Moles of NO=0.9459 mol HNO3×2 mol NO8 mol HNO3=0.2365 mol NO\text{Moles of } NO = 0.9459\,mol\,HNO_3 \times \frac{2\,mol\,NO}{8\,mol\,HNO_3} = 0.2365\,mol\,NO
      • Mass of NO=0.2365 mol×30.006 g mol−1=7.096 g≈7.10 g NO\text{Mass of } NO = 0.2365\,mol \times 30.006\,g\,mol^{-1} = 7.096\,g \approx 7.10\,g\,NO

Reaction Yield Analysis and Percent Yield Calculations

Percentage yield problem set and industrial chemical reactions

  • Theoretical vs. Actual Yield Relations:

    • Definition: Percentage yield is calculated using the formula:
      • Percentage Yield=Actual YieldTheoretical Yield×100%\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%
    • Reaction 1:
      • Theoretical yield: 38.0 g38.0\,g
      • Actual yield: 16.3 g16.3\,g
      • Percentage Yield=16.3 g38.0 g×100%=42.8%\text{Percentage Yield} = \frac{16.3\,g}{38.0\,g} \times 100\% = 42.8\%
    • Reaction 2:
      • Theoretical yield: 8.5 g8.5\,g
      • Actual yield: 111 mg=0.111 g111\,mg = 0.111\,g
      • Percentage Yield=0.111 g8.5 g×100%=1.3%\text{Percentage Yield} = \frac{0.111\,g}{8.5\,g} \times 100\% = 1.3\%
    • Reaction 3:
      • Theoretical yield: 3.3 metric tons=3300 kg3.3\,\text{metric tons} = 3300\,kg
      • Actual yield: 1277 kg1277\,kg
      • Percentage Yield=1277 kg3300 kg×100%=38.697%≈39%\text{Percentage Yield} = \frac{1277\,kg}{3300\,kg} \times 100\% = 38.697\% \approx 39\%
    • Reaction 4:
      • Theoretical yield: 40.0 g40.0\,g
      • Actual yield: 41.0 g41.0\,g
      • Percentage Yield=41.0 g40.0 g×100%=102.5%≈103%\text{Percentage Yield} = \frac{41.0\,g}{40.0\,g} \times 100\% = 102.5\% \approx 103\%
      • Caveat: Yields exceeding 100%100\% typically indicate the presence of retained solvent, wet product, or unreacted impurities in the collected sample.

  • Methanol Fuel Synthesis:

    • Industrial and automotive application: Methanol (CH3OHCH_3OH) is utilized in racing vehicles as a clean-burning fuel.
    • Balanced equation:
      • CO(g)+2 H2(g)→CH3OH(l)CO(g) + 2\,H_2(g) \rightarrow CH_3OH(l)
    • Reaction parameters: 4.8×103 g H24.8 \times 10^3\,g\,H_2 reacts with excess carbon monoxide (COCO) to produce an actual yield of 3.0×104 g CH3OH3.0 \times 10^4\,g\,CH_3OH.
    • Theoretical yield calculation:
      • Moles of H2=4.8×103 g2.016 g mol−1=2380.95 mol H2\text{Moles of } H_2 = \frac{4.8 \times 10^3\,g}{2.016\,g\,mol^{-1}} = 2380.95\,mol\,H_2
      • Theoretical moles of CH3OH=2380.95 mol H22=1190.48 mol CH3OH\text{Theoretical moles of } CH_3OH = \frac{2380.95\,mol\,H_2}{2} = 1190.48\,mol\,CH_3OH
      • Molar mass of CH3OH=12.011+4(1.008)+15.999=32.042 g mol−1\text{Molar mass of } CH_3OH = 12.011 + 4(1.008) + 15.999 = 32.042\,g\,mol^{-1}
      • Theoretical mass of CH3OH=1190.48 mol×32.042 g mol−1=3.815×104 g\text{Theoretical mass of } CH_3OH = 1190.48\,mol \times 32.042\,g\,mol^{-1} = 3.815 \times 10^4\,g
    • Percent yield calculation:
      • Percent Yield=3.0×104 g3.815×104 g×100%=78.6%≈79%\text{Percent Yield} = \frac{3.0 \times 10^4\,g}{3.815 \times 10^4\,g} \times 100\% = 78.6\% \approx 79\%

  • Synthesis of Carbon Disulfide from Methane and Elemental Sulfur:

    • Process condition: The reaction consistently exhibits a percent yield of 93%93\%.
    • Balanced chemical equation:
      • CH4(g)+4 S(g)→CS2(g)+2 H2S(g)CH_4(g) + 4\,S(g) \rightarrow CS_2(g) + 2\,H_2S(g)
    • Target actual yield: 83.2 g CS283.2\,g\,CS_2
    • Determination of required reactant mass:
      • Theoretical yield needed=83.2 g0.93=89.462 g CS2\text{Theoretical yield needed} = \frac{83.2\,g}{0.93} = 89.462\,g\,CS_2
      • Molar mass of CS2=12.011+2(32.065)=76.141 g mol−1\text{Molar mass of } CS_2 = 12.011 + 2(32.065) = 76.141\,g\,mol^{-1}
      • Theoretical moles of CS2=89.462 g76.141 g mol−1=1.1749 mol CS2\text{Theoretical moles of } CS_2 = \frac{89.462\,g}{76.141\,g\,mol^{-1}} = 1.1749\,mol\,CS_2
      • Moles of sulfur (S) required=1.1749 mol CS2×4=4.6998 mol S\text{Moles of sulfur (S) required} = 1.1749\,mol\,CS_2 \times 4 = 4.6998\,mol\,S
      • Molar mass of S=32.065 g mol−1\text{Molar mass of } S = 32.065\,g\,mol^{-1}
      • Mass of sulfur required=4.6998 mol×32.065 g mol−1=150.7 g\text{Mass of sulfur required} = 4.6998\,mol \times 32.065\,g\,mol^{-1} = 150.7\,g
    • Reported final value: 1.5×102 g1.5 \times 10^2\,g (1.5e2 g1.5e2\,g).

Acid-Base Titrations and Solution Stoichiometry

Acid-base titration calculations and solution stoichiometry

  • Titration of Solid Sodium Hydroxide with Hydrochloric Acid:

    • Balanced neutralization equation:
      • NaOH(aq)+HCl(aq)→H2O(l)+NaCl(aq)NaOH(aq) + HCl(aq) \rightarrow H_2O(l) + NaCl(aq)
    • Problem statement: Determine the volume in milliliters of 0.879 M HCl0.879\,M\,HCl required to titrate 1.30 g NaOH1.30\,g\,NaOH to the equivalence point.
    • Stoichiometric calculation:
      • Molar mass of NaOH=22.990+15.999+1.008=39.997 g mol−1\text{Molar mass of } NaOH = 22.990 + 15.999 + 1.008 = 39.997\,g\,mol^{-1}
      • Moles of NaOH=1.30 g39.997 g mol−1=0.03250 mol\text{Moles of } NaOH = \frac{1.30\,g}{39.997\,g\,mol^{-1}} = 0.03250\,mol
      • Moles of HCl required=0.03250 mol\text{Moles of } HCl \text{ required} = 0.03250\,mol
      • Volume of HCl=0.03250 mol0.879 mol L−1=0.03698 L=37.0 mL\text{Volume of } HCl = \frac{0.03250\,mol}{0.879\,mol\,L^{-1}} = 0.03698\,L = 37.0\,mL

  • Standardization of Nitric Acid via Barium Hydroxide:

    • Balanced neutralization equation:
      • Ba(OH)2(aq)+2 HNO3(aq)→Ba(NO3)2(aq)+2 H2O(l)Ba(OH)_2(aq) + 2\,HNO_3(aq) \rightarrow Ba(NO_3)_2(aq) + 2\,H_2O(l)
    • Problem statement: Determine the molarity of a nitric acid solution if 0.327 g0.327\,g of barium hydroxide is required to neutralize 20.00 mL20.00\,mL of nitric acid.
    • Stoichiometric calculation:
      • Molar mass of Ba(OH)2=137.327+2(15.999+1.008)=171.341 g mol−1\text{Molar mass of } Ba(OH)_2 = 137.327 + 2(15.999 + 1.008) = 171.341\,g\,mol^{-1}
      • Moles of Ba(OH)2=0.327 g171.341 g mol−1=0.0019085 mol\text{Moles of } Ba(OH)_2 = \frac{0.327\,g}{171.341\,g\,mol^{-1}} = 0.0019085\,mol
      • Moles of HNO3 neutralized=0.0019085 mol×2=0.003817 mol\text{Moles of } HNO_3 \text{ neutralized} = 0.0019085\,mol \times 2 = 0.003817\,mol
      • Molarity of HNO3=0.003817 mol0.02000 L=0.19085 M≈0.191 M\text{Molarity of } HNO_3 = \frac{0.003817\,mol}{0.02000\,L} = 0.19085\,M \approx 0.191\,M

  • Neutralization of Sulfuric Acid with Sodium Hydroxide:

    • Balanced neutralization equation:
      • 2 NaOH(aq)+H2SO4(aq)→Na2SO4(aq)+2 H2O(l)2\,NaOH(aq) + H_2SO_4(aq) \rightarrow Na_2SO_4(aq) + 2\,H_2O(l)
    • Problem statement: Determine the volume in milliliters of 0.502 M NaOH0.502\,M\,NaOH required to react completely with 25.0 mL25.0\,mL of 0.344 M H2SO40.344\,M\,H_2SO_4.
    • Stoichiometric calculation:
      • Moles of H2SO4=0.0250 L×0.344 mol L−1=0.00860 mol\text{Moles of } H_2SO_4 = 0.0250\,L \times 0.344\,mol\,L^{-1} = 0.00860\,mol
      • Moles of NaOH required=0.00860 mol×2=0.01720 mol\text{Moles of } NaOH \text{ required} = 0.00860\,mol \times 2 = 0.01720\,mol
      • Volume of NaOH=0.01720 mol0.502 mol L−1=0.03426 L=34.3 mL\text{Volume of } NaOH = \frac{0.01720\,mol}{0.502\,mol\,L^{-1}} = 0.03426\,L = 34.3\,mL

  • Titration and Standardization of Hydroiodic Acid with Calcium Hydroxide:

    • Balanced neutralization equation:
      • 2 HI(aq)+Ca(OH)2(aq)→CaI2(aq)+2 H2O(l)2\,HI(aq) + Ca(OH)_2(aq) \rightarrow CaI_2(aq) + 2\,H_2O(l)
    • Problem statement: An aqueous solution of hydroiodic acid (HIHI) is standardized against 0.140 M Ca(OH)20.140\,M\,Ca(OH)_2. Calculate the acid molarity if 10.5 mL10.5\,mL of base neutralizes 18.8 mL18.8\,mL of acid.
    • Stoichiometric calculation:
      • Moles of Ca(OH)2=0.0105 L×0.140 mol L−1=0.00147 mol\text{Moles of } Ca(OH)_2 = 0.0105\,L \times 0.140\,mol\,L^{-1} = 0.00147\,mol
      • Moles of HI neutralized=0.00147 mol×2=0.00294 mol\text{Moles of } HI \text{ neutralized} = 0.00147\,mol \times 2 = 0.00294\,mol
      • Molarity of HI=0.00294 mol0.0188 L=0.15638 M≈0.156 M\text{Molarity of } HI = \frac{0.00294\,mol}{0.0188\,L} = 0.15638\,M \approx 0.156\,M