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Química Geral · Atoms, moles and stoichiometry

Estequiometria e rendimento de reação

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O que você vai aprender

  • Balance chemical equations, including combustion reactions, by changing coefficients only.
  • Use the mole ratios of a balanced equation to convert between moles and masses of reactants and products.
  • Identify the limiting reactant and calculate the theoretical yield and the amount of excess reactant left over.
  • Calculate theoretical, actual and percent yields, and the overall yield of a multistep process.

Lição gratuita

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Ler a aula gratuita (6 partes)

Balancing equations

A balanced equation has the same number of atoms of each element on both sides, because atoms are neither created nor destroyed; total mass is therefore conserved, but the number of molecules need not be. Balance by changing coefficients, never subscripts: changing H₂O to H₂O₂ makes a different substance. For a combustion reaction, balance carbon first (one CO₂ per C), then hydrogen (one H₂O per two H), and oxygen last, because O₂ is the only species that contains nothing but oxygen. Propane (drawn): C₃H₈ + O₂ → CO₂ + H₂O gives 3CO₂ and 4H₂O, which hold 3 × 2 + 4 = 10 O atoms, so 5O₂: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. If oxygen comes out odd, as with C₂H₆ (2CO₂ + 3H₂O needs 7 O atoms, 3½O₂), double everything: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. If the fuel itself contains oxygen, subtract its O atoms before choosing the O₂ coefficient: C₂H₅OH gives 2CO₂ + 3H₂O (7 O), ethanol supplies 1, so 6 O come from 3O₂. Treat a polyatomic ion that survives unchanged (SO₄²⁻, PO₄³⁻) as a single unit. For a redox equation, first make the electrons lost equal the electrons gained, then balance the other atoms: if each Mn goes from +7 to +2 (gains 5 e⁻) and each Cl₂ forms from two Cl⁻ (loses 2 e⁻), then 2 Mn pair with 5 Cl₂.

Mole ratios and mass-to-mass conversions

The coefficients give mole ratios, not mass ratios. In C₃H₈ + 5O₂ → 3CO₂ + 4H₂O, one mole of propane consumes five moles of O₂ and forms three moles of CO₂. To go from a mass of one substance to a mass of another, always pass through moles: grams of A ÷ molar mass of A → moles of A; × (coefficient of B ÷ coefficient of A) → moles of B; × molar mass of B → grams of B. Example: burning 5.00 g of methane (16.04 g/mol, drawn), CH₄ + 2O₂ → CO₂ + 2H₂O, gives 5.00 ÷ 16.04 = 0.3117 mol CH₄ → 0.3117 mol CO₂ → 0.3117 × 44.01 = 13.7 g CO₂. The product mass is larger than the fuel mass because oxygen from the air is added.

Limiting and excess reactants

When the reactants are not in the exact ratio of the equation, one runs out first: the limiting reactant. It fixes how much product can form; the other is in excess and some is left over. Convert each reactant to moles and divide by its coefficient; the smallest result marks the limiting reactant, whatever its mass. Example: 5.00 g Na (22.99 g/mol) and 7.09 g Cl₂ (70.90 g/mol) react as 2Na + Cl₂ → 2NaCl. Na: 0.2175 mol ÷ 2 = 0.1087; Cl₂: 0.1000 mol ÷ 1 = 0.1000, so Cl₂ is limiting. NaCl formed = 2 × 0.1000 = 0.2000 mol × 58.44 g/mol = 11.69 g. Na used = 0.2000 mol × 22.99 = 4.60 g, so 5.00 − 4.60 = 0.40 g of Na is left over. Mass balances: 5.00 + 7.09 = 12.09 g = 11.69 g NaCl + 0.40 g Na.

Theoretical, actual and percent yield

The theoretical yield is the amount of product calculated from the limiting reactant, assuming complete reaction and no losses. The actual yield is what is measured. Percent yield = actual ÷ theoretical × 100. If the sodium chloride reaction above gave 10.5 g of NaCl, the percent yield would be 10.5 ÷ 11.69 × 100 = 89.8%. Actual yields fall short because reactions may not go to completion, side reactions form other products, and material is lost during transfer, filtration and purification. Only more of the limiting reactant raises the theoretical yield; adding more of the excess reactant does not. To work backwards from a target mass of product, first divide it by the fractional yield to get the theoretical amount you must plan for (30.0 g at 75.0% means planning for 30.0 ÷ 0.750 = 40.0 g), then convert that to moles of reactant.

Multistep reactions

When the product of one reaction is the reactant of the next, chain the mole ratios: multiply through every step's ratio from the starting material to the final product, and do not assume each step is 1 : 1. For yields, the fraction that survives is the product of the step yields, not their average and not the lowest one. Example: 0.500 mol of a starting material taken through two 1 : 1 steps with yields of 80.0% and 90.0% gives 0.500 × 0.800 × 0.900 = 0.360 mol of final product, an overall yield of 72.0%. A long sequence of good steps can end lower than a short sequence that contains one poor step, or higher, so compare overall yields by multiplying.

What to carry forward

Balance with coefficients, then always work in moles: convert masses to moles, use the coefficient ratio, and convert back. Divide each reactant's moles by its coefficient to find the limiting reactant; it alone sets the theoretical yield, and the excess left over is what was present minus what reacted. Percent yield is actual ÷ theoretical × 100, and the overall yield of a sequence is the product of the step yields.

Problemas de prática (40)

Praticar é grátis com uma conta.

Preveja o produto 4

Balancing equations

  • Making phosphoric acid from phosphate rockBalance an equation with polyatomic ions by treating ions that pass through unchanged as units.
  • Coefficients for reducing iron(III) oxideFind the smallest whole-number coefficients for the blast-furnace reduction of iron(III) oxide by carbon monoxide.

Percent yield and multistep reactions

  • What raises the theoretical yield?Predict which change alters the theoretical yield, as opposed to the actual yield or the rate.

Limiting and excess reactants

  • What is in the vessel afterwards?Predict the amounts of every substance present after a reaction with a limiting reactant goes to completion.

Classificar 2

Balancing equations

  • Which aluminium oxide equation is balanced?Recognise the correctly balanced equation for aluminium reacting with oxygen among equations with wrong coefficients or formulas.

Limiting and excess reactants

  • Which reactant runs out: zinc or HCl?Identify the limiting reactant from mole amounts when the coefficients are unequal.

Comparar 4

Mole ratios and mass stoichiometry

  • Carbon dioxide per gram of fuelRank fuels by the mass of CO₂ released per gram burned by combining mole ratios with molar masses.
  • Oxygen demand of four fuelsRank fuels by the moles of oxygen one mole of each needs for complete combustion, balancing each equation.

Percent yield and multistep reactions

  • Three good steps or two uneven ones?Compare the overall yields of two multistep routes by multiplying their step yields.

Limiting and excess reactants

  • A mixture with nothing left overCompare reactant mixtures to find the one in the exact stoichiometric mole ratio.

Calcular 19

Percent yield and multistep reactions

  • Expected mass at a known percent yieldCalculate the actual yield expected from a theoretical yield and a percent yield.
  • Percent yield when one reactant is in excessIdentify the limiting reactant from two masses and use it to calculate a percent yield.
  • Percent yield of quicklimeCalculate a percent yield starting from the mass of reactant rather than a given theoretical yield.
  • Nitric acid from ammonia in three stepsChain the mole ratios of three successive reactions to find the maximum mass of the final product.
  • Percent yield of magnesium oxideCalculate a percent yield from a given theoretical yield and a measured actual yield.
  • Sulfuric acid through three stepsCombine a theoretical yield with the step yields of a three-step sequence to find the mass of final product obtained.

Limiting and excess reactants

  • Aluminium chloride from aluminium and chlorineCalculate a theoretical yield when the reactant with the larger number of moles is the limiting one.
  • Theoretical yield of ammoniaFind the limiting reactant from two masses and calculate the theoretical yield of product in grams.
  • Oxygen left after burning hydrogenCalculate the moles of excess reactant remaining once the limiting reactant is used up.
  • Iron from a limited supply of carbon monoxideFind the limiting reactant when one reactant has the larger number of moles but also the larger coefficient, and calculate the product mass.
  • White phosphorus left after chlorinationCalculate the mass of excess reactant left over when the limiting reactant has a large coefficient.

Mole ratios and mass stoichiometry

  • Moles of ammonia from hydrogenConvert moles of one reactant to moles of product with the mole ratio from the balanced equation.
  • Oxygen needed to burn butaneBalance a combustion equation and use it to find the moles of oxygen consumed by a given amount of fuel.
  • Carbon dioxide from burning propaneCarry out a mass-to-mass calculation for a combustion reaction whose product coefficient is greater than one.
  • Aluminium for a thermite chargeCalculate the mass of one reactant needed to consume a given mass of the other.
  • Mass of water from hydrogenConvert a reactant mass to a product mass through moles for a reaction with a one-to-one mole ratio.

Balancing equations

  • Balancing octane combustionBalance a combustion equation that first gives a fractional O₂ coefficient and clear it to whole numbers.
  • Chlorine from permanganate and HClBalance a redox equation in which hydrochloric acid acts both as reducing agent and as a source of chloride ions.
  • Oxygen needed to burn propaneBalance the complete combustion of propane and read off the coefficient of oxygen.

Encontre o erro 2

Limiting and excess reactants

  • Choosing the limiting reactant by massDiagnose the error in choosing a limiting reactant by comparing masses instead of moles.

Balancing equations

  • Balancing by changing a formulaDiagnose why changing a subscript is not a valid way to balance an equation.

Interprete as evidências 2

Mole ratios and mass stoichiometry

  • Calcium carbonate content from mass lossInterpret the mass lost on heating a mixture as one gaseous product and use it to find the mass percent of the reacting component.
  • Reading a mole ratioRead the mole ratio between a reactant and a product from the coefficients of a balanced equation.

Trabalhe de trás para frente 4

Percent yield and multistep reactions

  • Ore needed for a kilogram of copperWork backwards from a target product mass through a percent yield, a mole ratio and an ore purity to the mass of ore required.

Balancing equations

  • Identify a hydrocarbon from its combustionWork backwards from the balanced combustion amounts to the molecular formula of a hydrocarbon.

Mole ratios and mass stoichiometry

  • Potassium chlorate for a target mass of oxygenBalance a decomposition and work backwards from a required product mass to the reactant mass.

Limiting and excess reactants

  • How much oxygen was there at the start?Work backwards from the final contents of a vessel to the initial amount of the excess reactant.

Planeje uma síntese 1

  • Order of balancing glucose combustionPlan the order in which to balance the elements of a combustion equation for a compound that contains oxygen.

Explique por quê 2

Balancing equations

  • What a balanced equation conservesExplain that a balanced equation conserves atoms of each element, and therefore mass, but not the number of molecules.

Percent yield and multistep reactions

  • Why the actual yield falls shortExplain a genuine cause of an actual yield below the theoretical yield.

Travou em estequiometria e rendimento de reação? 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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