Comprehensive Chem 121 Study Guide: Dimensional Analysis, Temperature Scales, Significant Figures, Matter Classification, and Solution Stoichiometry

Principles of Dimensional Analysis and Metric Conversions

  • Metric prefix memorization is required for all fundamental metric-to-metric conversions (e.g., converting grams to kilograms or milligrams to micrograms).
  • Tables containing material densities and English-to-metric conversion factors are provided on examinations; however, internal metric conversions must be known.
  • Proper labeling in conversion factors requires maintaining a explicit distinction between solutions (mixtures) and pure chemical substances (e.g., an NaOHNaOH solution component does not cancel with pure NaOHNaOH).
  • Dimensional analysis problems frequently contain extraneous conversion factors that are unnecessary for reaching the solution; mapping out a clear path prior to calculation prevents errors.

Standard Unit Conversions

  • Metric System Conversions:

    • Convert 25.025.0 milligrams into micrograms: 2.5×104 μg2.5 \times 10^4\,\mu\text{g}
    • Convert 36.9 cm36.9\,\text{cm} into millimeters: 369 mm369\,\text{mm}
    • Convert 6539 kg6539\,\text{kg} into megagrams: 6.539 megagrams6.539\,\text{megagrams}
    • Convert 1.50×103 picoseconds1.50 \times 10^3\,\text{picoseconds} into nanoseconds: 1.50 ns1.50\,\text{ns}

  • Dimensional Area and Volume Conversions:

    • Convert 1.54×105 mm31.54 \times 10^5\,\text{mm}^3 into m3\text{m}^3: 1.54×10−4 m31.54 \times 10^{-4}\,\text{m}^3
    • Convert 1.32×104 m21.32 \times 10^4\,\text{m}^2 into km2\text{km}^2: 1.34×10−2 km21.34 \times 10^{-2}\,\text{km}^2
    • Convert 325 cm2325\,\text{cm}^2 into mm2\text{mm}^2: 3.25×104 mm23.25 \times 10^4\,\text{mm}^2
    • Number of cubic centimeters in 1 cubic meter1\,\text{cubic meter}: 1 m3=1×106 cm31\,\text{m}^3 = 1 \times 10^6\,\text{cm}^3
    • Number of square inches in 1 square foot1\,\text{square foot}: 1 ft2=144 in21\,\text{ft}^2 = 144\,\text{in}^2

  • Distance and Spatial Unit Conversions:

    • Convert 2.3×104 meters2.3 \times 10^4\,\text{meters} into miles: 14 miles14\,\text{miles}
    • Convert 44 mm44\,\text{mm} into inches: 1.7 in1.7\,\text{in}
    • Convert 0.0017 km0.0017\,\text{km} into millimeters: 1.7×103 mm1.7 \times 10^3\,\text{mm}
    • Convert 57 m357\,\text{m}^3 into cubic centimeters: 57×106 cm357 \times 10^6\,\text{cm}^3 or 5.7×107 cm35.7 \times 10^7\,\text{cm}^3

Applied Factor-Label and Word Problems

  • Canine Feeding Cost Calculation:

    • Problem Parameters: A German shepherd eats 5 cups5\,\text{cups} of food per day. A pound of food contains 10.5 cups10.5\,\text{cups}. A 50.0 lb50.0\,\text{lb} bag costs $35.50\$35.50.
    • Objective: Determine the cost to feed the dog for 1 week1\,\text{week}.
    • Answer: $2\$2

  • Mass of an Individual Iron Atom:

    • Problem Parameters: One mole of iron (FeFe) contains 6.0×1023 atoms6.0 \times 10^{23}\,\text{atoms} and weighs 55.85 grams55.85\,\text{grams}.
    • Objective: Calculate the mass of a single atom of iron.
    • Answer: 9.3×10−23 g9.3 \times 10^{-23}\,\text{g}

  • Currency Exchange and Accommodation Budgeting:

    • Problem Parameters: An American college student in Paris has 550 German marks550\,\text{German marks} to exchange into French francs. Conversion rates: 1 franc=0.185 dollars1\,\text{franc} = 0.185\,\text{dollars}, and 0.49 dollars=1.41 German marks0.49\,\text{dollars} = 1.41\,\text{German marks}. Hotel cost is 145 francs/day145\,\text{francs/day}.
    • Objective: Determine if the student has sufficient funds to stay for two nights in the hotel.
    • Answer: Yes, she has enough money. The hotel will cost 154 marks154\,\text{marks}.

  • Hypothetical Unit System Dimensional Analysis:

    • Problem Parameters: Given equivalences: 7 greeps=50 sods7\,\text{greeps} = 50\,\text{sods}, 16 nuds=5 glors16\,\text{nuds} = 5\,\text{glors}, 1 glor=22 snaff1\,\text{glor} = 22\,\text{snaff}, 1 snaff=12 sods1\,\text{snaff} = 12\,\text{sods}, and 1 fream=21 nuds1\,\text{fream} = 21\,\text{nuds}.
    • Objective: Determine how many greeps are contained in 4 freams4\,\text{freams}.
    • Answer: 1×103 greeps1 \times 10^3\,\text{greeps}

  • Vaccine Serum Volume Evaluation:

    • Problem Parameters: A health center must vaccinate 1555 students1555\,\text{students}, with each dose requiring 10.0 mL10.0\,\text{mL} of serum. Available supply is 1.75 quarts1.75\,\text{quarts}.
    • Objective: Determine if the available volume is sufficient.
    • Answer: No, it will not be enough. The total requirement is 16.4 qt16.4\,\text{qt}.

  • Volume Determination from Mass and Density (Lead):

    • Objective: Calculate the volume of 2.5 lbs2.5\,\text{lbs} of lead.
    • Answer: 1.0×102 cm31.0 \times 10^2\,\text{cm}^3 (derived using lead density).

  • Multi-Proportional Fruit Counting Problem:

    • Problem Parameters: A bushel contains twice as many apples as oranges. For every 3 oranges3\,\text{oranges} there are 2 pears2\,\text{pears}. For each bundle of grapes there is 1 banana1\,\text{banana}. There are three times more bananas than pears. Each bundle of grapes contains 12 grapes12\,\text{grapes}. The bushel contains 6 apples6\,\text{apples}.
    • Objective: Find the total number of individual grapes in the bushel.
    • Answer: 7×101 grapes7 \times 10^1\,\text{grapes}

  • Mercury Atom Sample Volume:

    • Problem Parameters: One mole of mercury (HgHg) contains 6.0×1023 atoms6.0 \times 10^{23}\,\text{atoms} and has a mass of 201 grams201\,\text{grams}.
    • Objective: Calculate the volume of a sample containing 2.0×1024 Hg atoms2.0 \times 10^{24}\,\text{Hg atoms}.
    • Answer: 49 cm349\,\text{cm}^3 (derived using mercury density).

  • Gold Mass Calculation:

    • Objective: Find the mass in pounds of a sample of gold having a volume of 5.60\,\text{in}^3$.\n - Answer: 3.91\,\text{lbs} (derived using gold density).\n\n- **Water Molecule Volume Calculation**:\n - Problem Parameters: One mole of H_2Ocontainscontains6.0 \times 10^{23}\,\text{molecules}andhasamassofand has a mass of18.0\,\text{grams}.\n - Objective: Find the volume in liters of a sample containing 3.0 \times 10^{24}\,\text{molecules}ofofH_2O$.
    • Answer: 0.090 L0.090\,\text{L}

  • Saline Solution Volume Calculation:

    • Problem Parameters: A solution of table salt (NaClNaCl) in water contains 23.0 g NaCl23.0\,\text{g NaCl} per 100 g solution100\,\text{g solution}. Solution density is 1.19 g solution/1 mL solution1.19\,\text{g solution} / 1\,\text{mL solution}. Molar mass of NaClNaCl is 58.45 g/mol58.45\,\text{g/mol}.
    • Objective: Determine the solution volume containing 0.55 moles0.55\,\text{moles} of NaCl$.\n - Answer: 1.2 \times 10^2\,\text{mL}\n\n- **Agricultural Area Conversion**:\n - Problem Parameters: An average US farm occupies 435\,\text{acres}.Equivalences:. Equivalences:1\,\text{km}^2 = 247\,\text{acres},,1\,\text{km} = 0.6214\,\text{miles}.\n - Objective: Calculate farm area in square miles.\n - Answer: 0.680\,\text{mil}^2\n\n- **Vineyard Soil Supplement Mass**:\n - Problem Parameters: A vineyard occupies 145\,\text{acres}.Soilsupplementapplicationrateis. Soil supplement application rate is5.50\,\text{g}perper1\,\text{m}^2.Equivalence:. Equivalence:1\,\text{km}^2 = 247\,\text{acres}.\n - Objective: Determine total kilograms of soil supplement required for the entire vineyard.\n - Answer: 3.23 \times 10^3\,\text{kg}\n\n- **International Fuel Price Conversion**:\n - Problem Parameters: Gas in France costs 1.42\,\text{euros}perliter.Equivalences:per liter. Equivalences:1\,\text{euro} = 1.10\,\text{US \$},,1.000\,\text{L} = 1.0567\,\text{quart}.\n - Objective: Convert price into US dollars per gallon.\n - Answer: \$5.91 / \text{gallon}\n\n- **Density Unit Conversion**:\n - Problem Parameters: Density of Feisis7.87\,\text{g/cm}^3.Equivalences:. Equivalences:1\,\text{lb} = 453.6\,\text{g},,1\,\text{in} = 2.54\,\text{cm}.\n - Objective: Convert density into \text{lb/in}^3$.
    • Answer: 0.284 lb/in30.284\,\text{lb/in}^3

  • Running Speed Conversion:

    • Problem Parameters: The men's 100 meter dash world record is 9.58 seconds9.58\,\text{seconds}. Equivalence: 1 km=0.6214 miles1\,\text{km} = 0.6214\,\text{miles}.
    • Objective: Calculate runner speed in miles per hour.
    • Answer: 23.4 mil/hr23.4\,\text{mil/hr}

Advanced Multi-Step Conversions and Solution Mapping

  • Mass of Lead Sample:

    • Problem: Find the mass in pounds of 1.50×103 in31.50 \times 10^3\,\text{in}^3 of lead.
    • Solution Map: in3→cm3→g→lb\text{in}^3 \rightarrow \text{cm}^3 \rightarrow \text{g} \rightarrow \text{lb}
    • Answer: 615 lbs615\,\text{lbs}

  • Medication Daily Dose Cost Analysis:

    • Problem: Daily recommended dose is 1.5×10−3 g medicine1.5 \times 10^{-3}\,\text{g medicine} per 1 kg body mass1\,\text{kg body mass}. Cost is $0.25\$0.25 per 15 mg15\,\text{mg}. Find 1-day supply cost for a 175 lb175\,\text{lb} individual.
    • Solution Map: \text{lb body} \rightarrow \text{g body} \rightarrow \text{kg body} \n\rightarrow \text{g med} \rightarrow \text{mg med} \rightarrow \text{cents} \rightarrow \
    • Answer: $2.0\$2.0

  • Lightbulb Electricity Cost Calculation:

    • Problem: A 75 W75\,\text{W} bulb requires 75 J/s75\,\text{J/s}. Electricity costs 13 cents13\,\text{cents} per kilowatt-hour. Equivalences: 1055 J/BTU1055\,\text{J/BTU}, 252 cal/BTU252\,\text{cal/BTU}, 1 kWh=3.61×103 kJ1\,\text{kWh} = 3.61 \times 10^3\,\text{kJ}. Find the cost in dollars to run the bulb continuously for 3.0 days3.0\,\text{days}.
    • Solution Map: \text{days} \rightarrow \text{hrs} \rightarrow \text{min} \rightarrow \text{sec} \rightarrow \text{joules} \rightarrow \text{kJ} \rightarrow \text{kilowatt hr} \rightarrow \text{cents} \rightarrow \
    • Answer: $0.70\$0.70

  • Phosphoric Acid (H3PO4H_3PO_4) Mass from Atom Count:

    • Problem: Each H3PO4H_3PO_4 molecule contains 3 atoms H3\,\text{atoms H}. Molar mass H=1.0 g/molH = 1.0\,\text{g/mol}, H3PO4=98.0 g/molH_3PO_4 = 98.0\,\text{g/mol}. Avogadro's number: 6.0×10236.0 \times 10^{23}. Find mass in kilograms of H3PO4H_3PO_4 containing 5.0×1025 H atoms5.0 \times 10^{25}\,\text{H atoms}.
    • Solution Map: H atoms→molecules H3PO4→mole H3PO4→g H3PO4→kg H3PO4\text{H atoms} \rightarrow \text{molecules } H_3PO_4 \rightarrow \text{mole } H_3PO_4 \rightarrow \text{g } H_3PO_4 \rightarrow \text{kg } H_3PO_4
    • Answer: 2.7 kg2.7\,\text{kg}

Temperature Conversions and Custom Temperature Scales

  • Standard Temperature Conversions:

    • Convert 37.5∘C37.5^\circ\text{C} into Kelvin: 310.7 K310.7\,\text{K}
    • Convert 253 K253\,\text{K} into Celsius: −20.∘C-20.^\circ\text{C}
    • Exothermic reaction temperature change ΔT=11.9∘C\Delta T = 11.9^\circ\text{C} converted to Fahrenheit change: 21.4∘F21.4^\circ\text{F}
    • Atmospheric temperature drop ΔT=25.6∘F\Delta T = 25.6^\circ\text{F} converted to Celsius change: 14.2∘C14.2^\circ\text{C}
    • Human body temperature 98.6∘F98.6^\circ\text{F} converted to Celsius: 37.0∘C37.0^\circ\text{C}
    • Convert 125∘C125^\circ\text{C} into Fahrenheit: 257∘F257^\circ\text{F}
    • Atmospheric temperature 57∘F57^\circ\text{F} converted to Celsius: 13.(9)∘C13.(9)^\circ\text{C}
    • Atmospheric temperature 21∘C21^\circ\text{C} converted to Fahrenheit: 69.(8)∘F69.(8)^\circ\text{F}

  • Hypothetical Temperature Scale (∘H^\circ\text{H}) Analysis:

    • Scale Definition: Water boils at 75∘H75^\circ\text{H} and freezes at −55∘H-55^\circ\text{H}. Total range between freezing and boiling points is 75−(−55)=130∘H75 - (-55) = 130^\circ\text{H}.
    • Degree Unit Size Comparison (∘H^\circ\text{H} vs ∘F^\circ\text{F}): 1∘H1^\circ\text{H} is larger than 1∘F1^\circ\text{F}. The difference between the freezing and boiling points of water is 130∘H130^\circ\text{H}, which is less than the 180∘F180^\circ\text{F} difference on the Fahrenheit scale. Because the Fahrenheit scale has a higher number of degree divisions across the same physical temperature span, each individual ∘F^\circ\text{F} unit is smaller than 1∘H1^\circ\text{H}.
    • Degree Unit Size Comparison (∘H^\circ\text{H} vs ∘C^\circ\text{C}): 1∘H1^\circ\text{H} is smaller than 1∘C1^\circ\text{C} (since Celsius contains 100∘C100^\circ\text{C} divisions across the water phase transition span compared to 130∘H130^\circ\text{H} divisions).

  • Comparative Temperature Change Calculations:

    • If water ΔT=5∘C\Delta T = 5^\circ\text{C}, the change in Fahrenheit is 9∘F9^\circ\text{F}.
    • If water ΔT=15∘H\Delta T = 15^\circ\text{H}, the change in Celsius is:     15∘H×(100∘C130∘H)=11.5∘C15^\circ\text{H} \times \left(\frac{100^\circ\text{C}}{130^\circ\text{H}}\right) = 11.5^\circ\text{C}
    • If water ΔT=10∘F\Delta T = 10^\circ\text{F}, the change in Hypothetical scale is:     10∘F×(130∘H180∘F)=7.2∘H10^\circ\text{F} \times \left(\frac{130^\circ\text{H}}{180^\circ\text{F}}\right) = 7.2^\circ\text{H}
    • Atmospheric temperature change ΔT=10∘F\Delta T = 10^\circ\text{F} converted to Celsius: 5.5(6)∘C5.5(6)^\circ\text{C}

Significant Figures Rules and Mathematical Operations

  • Significant Figure Identification:

    • 25.025.0: 3 Significant Figures
    • 0.00630.0063: 2 Significant Figures
    • 450.0450.0: 4 Significant Figures
    • 0.50030.5003: 4 Significant Figures
    • 0.003500.00350: 3 Significant Figures
    • 400400: Ambiguous (lacks an explicit decimal point)
    • 400.400.: 3 Significant Figures (explicit decimal point preserves precision)

  • Calculations and Precision Rules:

    • Addition/Subtraction: Result is limited by the least precise decimal place.
    • Multiplication/Division: Result is limited by the lowest total number of significant figures.
    • Calculation Exercises:
    • 13.5712+0.020=13.59113.5712 + 0.020 = 13.591
    • 94.8753−94.625=0.25094.8753 - 94.625 = 0.250
    • (16.135)(0.00430)=0.0694(16.135)(0.00430) = 0.0694
    • (2.200×103)(6.0×10−6)=0.013(2.200 \times 10^3)(6.0 \times 10^{-6}) = 0.013
    • 451×52=2.3×104451 \times 52 = 2.3 \times 10^4
    • (8.6×102)+124=9.8×102(8.6 \times 10^2) + 124 = 9.8 \times 10^2
    • 9.42+8.750.002750=6.607×103\frac{9.42 + 8.75}{0.002750} = 6.607 \times 10^3
    • (9.734−9.61)(1.50×102)=19(9.734 - 9.61)(1.50 \times 10^2) = 19
    • (82.5×1.76)+7.4=153(82.5 \times 1.76) + 7.4 = 153
    • 96.540.1326−613.52=114.5\frac{96.54}{0.1326} - 613.52 = 114.5
    • 8.73+7.5701.516×10−4=1.075×105\frac{8.73 + 7.570}{1.516 \times 10^{-4}} = 1.075 \times 10^5

Classification and Properties of Matter

  • Fixed Composition vs. Variable Composition:

    • Pure substances possess fixed compositions (specifically elements and compounds).
    • Pure substance examples include graphite (pure carbon) and glucose (C6H12O6C_6H_{12}O_6).
    • Propane (C3H8C_3H_8) is classified as a pure substance because it has a fixed chemical composition consisting of exactly 3 carbon atoms for every 8 hydrogen atoms. In contrast, mixtures like Kool-Aid have variable compositions due to varying sugar concentrations.
    • Uniform composition refers to having the exact same percent composition throughout the substance (e.g., identical concentration of sugar in every portion of Kool-Aid from a given batch).
    • Homogeneous mixtures possess a variable and uniform composition.

  • Classification Categorization of Substances:

    • Categories: i. homogeneous, ii. solution, iii. pure substance, iv. compound, v. element, vi. mixture, vii. heterogeneous
    • Nitrogen gas: Pure substance, Element
    • Hot coffee: Homogeneous, Solution, Mixture
    • Vinegar: Homogeneous, Solution, Mixture
    • Distilled water: Pure substance, Compound
    • Table salt (NaClNaCl): Pure substance, Compound
    • Oil and water: Mixture, Heterogeneous
    • Gatorade: Homogeneous, Solution, Mixture
    • Sugar (C12H22O11C_{12}H_{22}O_{11}): Pure substance, Compound
    • Aluminum: Pure substance, Element
    • Sandstone: Mixture, Heterogeneous
    • Orange juice without pulp: Homogeneous, Solution, Mixture
    • Orange juice with pulp: Solution, Mixture, Heterogeneous
    • Wine: Homogeneous, Solution, Mixture
    • Beef stew: Mixture, Heterogeneous
    • Iron rod: Pure substance, Compound
    • Carbon monoxide (COCO): Pure substance, Compound

Quantitative Solution Stoichiometry and Multi-Step Systems

  • Sulfuric Acid (H2SO4H_2SO_4) and Sodium Hydroxide (NaOHNaOH) Reaction System:

    • System Constants:
    • H2SO4H_2SO_4 Molar Mass: 98.0 g pure H2SO4/1 mole H2SO498.0\,\text{g pure } H_2SO_4 / 1\,\text{mole } H_2SO_4
    • NaOHNaOH Molar Mass: 40.0 g pure NaOH/1 mole NaOH40.0\,\text{g pure } NaOH / 1\,\text{mole } NaOH
    • Avogadro Constant: 6.0×1023 molecules/mole6.0 \times 10^{23}\,\text{molecules/mole}
    • H2SO4H_2SO_4 Solution Concentration: 1.5 moles pure H2SO4/1 L solution1.5\,\text{moles pure } H_2SO_4 / 1\,\text{L solution}
    • NaOHNaOH Solution Concentration: 23.0 g pure NaOH/100.0 g solution23.0\,\text{g pure } NaOH / 100.0\,\text{g solution}
    • NaOHNaOH Solution Density: 1.17 g solution/1 mL solution1.17\,\text{g solution} / 1\,\text{mL solution}
    • Reaction Stoichiometry: 1 mole pure H2SO41\,\text{mole pure } H_2SO_4 reacts with 2 moles pure NaOH2\,\text{moles pure } NaOH
    • Sub-Problem Calculations:
    • Mass of Pure Acid in Solution Volume: Find grams of pure H2SO4H_2SO_4 required to prepare 75 mL75\,\text{mL} of H2SO4H_2SO_4 solution.
      • Map: mL soln→L soln→mol pure H2SO4→grams pure H2SO4\text{mL soln} \rightarrow \text{L soln} \rightarrow \text{mol pure } H_2SO_4 \rightarrow \text{grams pure } H_2SO_4
      • Answer: 11 g11\,\text{g}
    • Volume of Base Solution for Given Moles: Find mL of NaOHNaOH solution containing 2.75 moles2.75\,\text{moles} of pure NaOHNaOH
      • Map: moles pure NaOH→grams pure NaOH→grams solution→mL solution\text{moles pure } NaOH \rightarrow \text{grams pure } NaOH \rightarrow \text{grams solution} \rightarrow \text{mL solution}
      • Answer: 409 mL409\,\text{mL}
    • Molecular Count from Mass: Find number of H2SO4H_2SO_4 molecules contained in 35 g35\,\text{g} of H2SO4H_2SO_4
      • Map: g H2SO4→mole H2SO4→molecules H2SO4\text{g } H_2SO_4 \rightarrow \text{mole } H_2SO_4 \rightarrow \text{molecules } H_2SO_4
      • Answer: 2.1×1023 molecules2.1 \times 10^{23}\,\text{molecules}
    • Stoichiometric Reactant Mass: Find grams of H2SO4H_2SO_4 required to completely react with 100.0 g100.0\,\text{g} of pure NaOHNaOH
      • Map: g pure NaOH→mole NaOH→mole pure H2SO4→grams H2SO4\text{g pure } NaOH \rightarrow \text{mole } NaOH \rightarrow \text{mole pure } H_2SO_4 \rightarrow \text{grams } H_2SO_4
      • Answer: 123 grams123\,\text{grams}
    • Volumetric Solution Reacting Stoichiometry: Find mL of NaOHNaOH solution required to completely react with 50.0 mL50.0\,\text{mL} of H2SO4H_2SO_4 solution.
      • Map: mL H2SO4 soln→L H2SO4 soln→mol pure H2SO4→mol pure NaOH→grams pure NaOH→g NaOH soln→mL NaOH soln\text{mL } H_2SO_4 \text{ soln} \rightarrow \text{L } H_2SO_4 \text{ soln} \rightarrow \text{mol pure } H_2SO_4 \rightarrow \text{mol pure } NaOH \rightarrow \text{grams pure } NaOH \rightarrow \text{g } NaOH \text{ soln} \rightarrow \text{mL } NaOH \text{ soln}
      • Answer: 22.3 mL22.3\,\text{mL}

  • Complex Acid-Base Solution System (H2SO4H_2SO_4 and Al(OH)3Al(OH)_3):

    • System Parameters:
    • Aqueous H2SO4H_2SO_4 Solution: 95 g pure H2SO4/100 g solution95\,\text{g pure } H_2SO_4 / 100\,\text{g solution}, density = 1.84 g solution/1 mL solution1.84\,\text{g solution} / 1\,\text{mL solution}, concentration = 36 equivalents pure H2SO4/1 L solution36\,\text{equivalents pure } H_2SO_4 / 1\,\text{L solution}, molar mass = 98.09 g H2SO4/mole98.09\,\text{g } H_2SO_4 / \text{mole}.
    • Aqueous Al(OH)3Al(OH)_3 Solution: 34 g pure Al(OH)3/100 g solution34\,\text{g pure } Al(OH)_3 / 100\,\text{g solution}, normal concentration = 16 equivalents pure Al(OH)3/1 L solution16\,\text{equivalents pure } Al(OH)_3 / 1\,\text{L solution}, mass per equivalent = 26.00 g/equivalent26.00\,\text{g/equivalent}, molar mass = 78.01 g Al(OH)3/mole78.01\,\text{g } Al(OH)_3 / \text{mole}.
    • Reaction Equivalence: 1 equivalent of Al(OH)31\,\text{equivalent of } Al(OH)_3 reacts with 1 equivalent of H2SO41\,\text{equivalent of } H_2SO_4
    • Calculation:
    • Determine volume in mL of Al(OH)3Al(OH)_3 solution needed to react completely with 25 g25\,\text{g} of H2SO4H_2SO_4 solution.
    • Map: g H2SO4 solution→mL H2SO4 solution→L H2SO4 solution→equiv H2SO4→equiv Al(OH)3→L Al(OH)3 solution\text{g } H_2SO_4 \text{ solution} \rightarrow \text{mL } H_2SO_4 \text{ solution} \rightarrow \text{L } H_2SO_4 \text{ solution} \rightarrow \text{equiv } H_2SO_4 \rightarrow \text{equiv } Al(OH)_3 \rightarrow \text{L } Al(OH)_3 \text{ solution}
    • Answer: 30.6 mL30.6\,\text{mL}