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Química Geral · Exam review

Revisão para a prova: estequiometria e reações

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O que você vai aprender

  • Apply significant-figure and unit-conversion rules in multi-step calculations, including prefixes and squared or cubed units.
  • Convert among mass, moles and particles, and deduce empirical, molecular and hydrate formulas from analytical data.
  • Find the limiting reactant and calculate theoretical yield, percent yield, excess reactant left over and the starting amounts a target needs.
  • Solve solution stoichiometry problems: ion concentrations, precipitation on mixing, and acid–base and redox titrations that use aliquots.
  • Write net ionic equations, classify reactions, and assign oxidation states to identify oxidizing and reducing agents.

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Measurements, units and significant figures

Multiplication and division: the result keeps as many significant figures as the least precise factor. Addition and subtraction: the result keeps as many decimal places as the least precise term. Exact numbers (counted objects, defined conversions such as 1 dL = 100 mL or 1 µm = 10⁻⁴ cm, coefficients in equations) never limit precision. Carry one or two extra digits through a calculation and round once at the end. Worked example: a beaker weighs 40.318 g empty and 48.27 g after 10.00 mL of a liquid is added. The liquid's mass is 48.27 − 40.318 = 7.952 g, known only to the hundredths place, so 7.95 g (3 significant figures); the density is 7.952 g ÷ 10.00 mL = 0.7952, reported as 0.795 g/mL. Squared and cubed units take the conversion factor squared or cubed: 1 m² = 10⁴ cm², 1 cm³ = 1 mL, 1 L = 10³ cm³. Exam trap: a dL is 100 mL, not 10 mL and not a litre.

Moles, formulas and hydrates

n = m/M turns mass into moles and N = n × Nᴀ (6.022 × 10²³ mol⁻¹) turns moles into particles; multiply by a subscript to count atoms of one element (a mole of C₆H₁₄ holds 14 mol of H atoms). Empirical formula from mass percent: take 100 g, convert each element's mass to moles, divide by the smallest, and multiply through if a ratio ends in .5 (×2) or in .33 or .67 (×3). Molecular formula = (molar mass ÷ empirical-formula mass) × empirical formula. Combustion analysis: all C ends up in CO₂ (n_C = n_CO₂), all H in H₂O (n_H = 2 n_H₂O), and a third element such as N or O is found by difference in mass. Hydrates: heating drives off the water. Worked example: 2.497 g of CuSO₄·xH₂O leaves 1.596 g of CuSO₄ (159.61 g/mol, drawn), which is 0.01000 mol; the water lost is 0.901 g ÷ 18.02 g/mol = 0.0500 mol, so x = 0.0500 ÷ 0.01000 = 5. Use the hydrate's own molar mass whenever you weigh the hydrate.

Limiting reactant and yield

Convert every reactant to moles and divide by its coefficient; the smallest quotient marks the limiting reactant. Never compare raw masses, and never compare raw moles when the coefficients differ. Worked example: N₂ + 3 H₂ → 2 NH₃ with 2.0 mol N₂ and 4.5 mol H₂. N₂ gives 2.0 ÷ 1 = 2.0 and H₂ gives 4.5 ÷ 3 = 1.5, so H₂ is limiting although there are more moles of it. NH₃ formed = 4.5 × 2/3 = 3.0 mol; N₂ used = 1.5 mol, so 0.5 mol N₂ is left over. Percent yield = actual ÷ theoretical × 100. To plan for a target, divide the target by the fractional yield (a 72 % yield needs 1/0.72 times the theoretical amount of reactant). In a sequence of steps, chain the mole ratios and multiply the step yields. A percent yield above 100 % signals an impure or wet product, never extra product from the excess reactant.

Solution stoichiometry and titrations

n = c × V; with volumes in mL, mL × mol/L = mmol. A strong electrolyte gives each ion at (salt concentration × subscript). When solutions are mixed, add up moles, let any precipitation run to completion with the limiting ion, and divide what is left by the total volume. In a titration the balanced equation supplies the mole ratio: 20.00 mL of 0.1000 M NaOH (2.000 mmol) neutralizes 1.000 mmol of diprotic H₂SO₄ (drawn), so a 10.00 mL sample is 0.1000 M. If a stock was diluted and only an aliquot titrated, scale back up by (total volume ÷ aliquot volume) and then undo the dilution. Redox titrations work the same way with the redox equation: MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O means 5 mol Fe²⁺ per mol MnO₄⁻. Error analysis: anything that makes the measured titre larger than the true one (diluted titrant, a burette bubble that escapes, an overshot end point) makes the calculated analyte concentration too high; diluting the analyte aliquot before it is delivered makes it too low.

Net ionic equations and reaction types

Split into ions only the strong electrolytes that are dissolved: strong acids (HCl, HBr, HI, HNO₃, HClO₄, H₂SO₄), strong bases (group 1 hydroxides, Ba(OH)₂) and soluble salts. Keep solids, weak acids and bases (CH₃COOH, HF, NH₃), gases and water intact, cancel the spectator ions and check both atoms and charge. Solubility rules: salts of Na⁺, K⁺, NH₄⁺ and NO₃⁻ are soluble; chlorides, bromides and iodides are soluble except with Ag⁺, Pb²⁺ and Hg₂²⁺; sulfates are soluble except with Ba²⁺, Sr²⁺ and Pb²⁺ (Ca²⁺ and Ag⁺ sulfates only slightly); hydroxides, carbonates, phosphates and sulfides are mostly insoluble except with group 1 cations and NH₄⁺, and Ba(OH)₂ is also soluble (Ca(OH)₂ and Sr(OH)₂ only slightly). Example: BaCl₂(aq) + Na₂SO₄(aq) gives the net ionic equation Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) (drawn), with Na⁺ and Cl⁻ as spectators; HF(aq) + OH⁻(aq) → F⁻(aq) + H₂O(l) keeps the weak acid whole. Reaction types: precipitation, acid–base neutralization, gas-forming (carbonates and hydrogen carbonates give CO₂ + H₂O, sulfites give SO₂), and redox, which includes combustion, single replacement and any combination or decomposition in which an element changes oxidation state.

Oxidation states and redox

Rules in priority order: an element is 0; a monatomic ion has its charge; F is −1; H is +1 with nonmetals; O is −2 except in peroxides (−1); the oxidation states sum to the charge of the species. In permanganate (drawn), Mn + 4(−2) = −1 gives Mn +7. Oxidation is an increase in oxidation state (loss of electrons); reduction is a decrease (gain). The species reduced is the oxidizing agent; the species oxidized is the reducing agent. An element that is both oxidized and reduced in one reaction disproportionates. Balancing in acid: split into half-reactions, balance the element, then O with H₂O, H with H⁺ and charge with electrons; multiply so the electrons cancel, add, and cancel H⁺ and H₂O that appear on both sides. In any redox reaction, moles of electrons lost = moles of electrons gained, so titration data can reveal how many electrons each oxidant accepts and therefore its product.

What to carry forward

Work in moles and let the balanced equation supply every ratio. Round once, at the end, to the precision the data allow. Divide moles by coefficients to find the limiting reactant, scale titration aliquots back to the whole sample, write only dissolved strong electrolytes as ions, and track electrons through oxidation states.

Problemas de prática (40)

Praticar é grátis com uma conta.

Preveja o produto 4

Limiting reactant and yield

  • Balance the combustion of glycerolBalance the combustion of a fuel that already contains oxygen, using the smallest whole-number coefficients.

Net ionic equations and redox

  • Net ionic equation for a rust-brown precipitateWrite the net ionic equation of a precipitation reaction by removing spectator ions and balancing charge.
  • Net ionic equation for an antacidWrite the net ionic equation for a strong acid reacting with an insoluble base, keeping the solid intact.
  • Net ionic equation: vinegar and baking sodaWrite the net ionic equation for a gas-forming reaction of a weak acid with hydrogen carbonate.

Classificar 3

Limiting reactant and yield

  • Which rocket propellant runs out?Identify the limiting reactant when the reactant present in larger amount is the one used up.

Net ionic equations and redox

  • Find the oxidizing agentIdentify the oxidizing agent in a redox equation from the changes in oxidation state.
  • Classify the decomposition of hydrogen peroxideRecognize that a decomposition can also be a redox reaction by tracking the oxidation state of oxygen.

Comparar 3

Solutions and titrations

  • Rank acid samples by the base they consumeRank acid samples by the amount of base needed for complete neutralization, accounting for volume, concentration and the number of acidic hydrogens.

Net ionic equations and redox

  • Rank nitrogen species by oxidation stateAssign oxidation states to nitrogen across hydrides, the element, oxides, oxoacids and oxoanions, and rank them.

Measurement, moles and formulas

  • Rank samples by oxygen atomsRank samples given as masses, moles and molecule counts by the number of atoms of one element they contain.

Calcular 20

Limiting reactant and yield

  • Nitrogen from an airbag inflatorChain the mole ratios of two consecutive reactions to find the total amount of a product formed in both.
  • Percent yield of aspirinFind the limiting reactant from a mass and a liquid volume, then calculate a percent yield.
  • Carbon disulfide left after burningCalculate the mass of excess reactant remaining after the limiting reactant is used up.
  • Carbon dioxide from burning ethanolConvert a mass of fuel to the mass of a combustion product through the mole ratio of a balanced equation.
  • Titanium from titanium(IV) chlorideFind the theoretical yield of a metal from two reactant masses when the lighter reactant is limiting.

Measurement, moles and formulas

  • Thickness of aluminium foilCombine density, area and a prefix conversion to find a length too small to measure with a ruler.
  • Blood glucose from mg/dL to mmol/LConvert a clinical mass concentration in mg/dL into a molar concentration in mmol/L.
  • Copper in a sample of the blue hydrateFind the mass of one element in a weighed sample of a hydrate using the hydrate's molar mass.
  • Density from a mass found by differenceCombine the addition and multiplication rules for significant figures in a density measured by mass difference.
  • Hydrogen atoms in a volume of hexaneChain density, molar mass, a subscript and Avogadro's number to count atoms of one element in a measured volume of liquid.
  • Empirical formula of an orange saltDerive an empirical formula from mass percentages when the mole ratio comes out as a half-integer.

Net ionic equations and redox

  • Hypochlorite in bleach by iodometric titrationChain two redox equations and a dilution to find an oxidant's concentration from a thiosulfate titre.
  • Balance the dichromate–ethanol reactionBalance a redox equation in acid by half-reactions and cancel the species that appear on both sides.
  • Oxidation state of chlorine in pool chlorineAssign the oxidation state of an element in a salt with a polyatomic anion from the cation charge and the oxygen rule.

Solutions and titrations

  • Composition of a carbonate–bicarbonate mixtureSet up and solve two simultaneous mass and mole balances to find the composition of a two-component mixture from one titration.
  • Nitrate ion in iron(III) nitrate solutionCalculate the concentration of one ion in a solution made from a weighed salt that releases several of that ion.
  • Iron in a supplement tablet by permanganate titrationUse a redox titration of an aliquot to find the mass of iron in a whole tablet.
  • Mass of lead(II) iodide from mixing two solutionsFind the limiting ion when two salt solutions are mixed and calculate the mass of precipitate.
  • Titre for a diprotic acidPredict the titrant volume needed when the analyte acid gives up two protons.
  • Ion left over after silver chromate precipitatesFind which precipitating ion is in excess after mixing and its concentration in the combined volume.

Encontre o erro 2

Limiting reactant and yield

  • Find the error in a limiting-reactant answerDiagnose a limiting-reactant choice made by comparing moles without dividing by the coefficients.

Solutions and titrations

  • Which titration error gives a low result?Trace how each common titration error changes the measured titre and the calculated analyte concentration.

Interprete as evidências 2

Measurement, moles and formulas

  • Molecular formula from combustion dataUse combustion masses, nitrogen by difference and an approximate molar mass to choose a molecular formula.

Net ionic equations and redox

  • What does permanganate become?Use the electron balance from titration data to find the oxidation state, and so the product, of a reduced oxidant.

Trabalhe de trás para frente 3

Measurement, moles and formulas

  • Which carbonate was heated?Identify a metal from the mass lost when its carbonate decomposes to the oxide.
  • Water of crystallization in a nickel saltWork back from the mass lost on heating a hydrate to its number of waters of crystallization.

Limiting reactant and yield

  • Ammonia needed for a target mass of ureaWork backwards from a target product mass and a percent yield to the reactant mass required.

Planeje uma síntese 2

Solutions and titrations

  • How much borax to weighPlan the mass of a hydrated primary standard that will give a convenient titre.
  • Prepare a standard dichromate solutionChoose the mass and glassware procedure that gives a solution of exactly known molarity.

Explique por quê 1

  • A yield above 100 %Explain why a measured percent yield above 100 % points to an impure or wet product.

Travou em revisão para a prova: estequiometria e reações? Escaneie o problema.

Tire uma foto da reação da sua lista ou do livro e desenhe o mecanismo com cada seta corrigida. 1 scan grátis agora, sem cadastro; mais 1 com uma conta gratuita.

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