3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A mixture of 3.00 g of H₂(g) and 16.0 g of O₂(g) reacts according to the equation 2 H₂(g) + O₂(g) → 2 H₂O(l). What is the maximum mass of H₂O that can form?
Answer and reasoning
A19.0 g A student who thinks every reactant is used up completely picks this, adding 3.00 g and 16.0 g. Only 2.0 g of the H₂ reacts; about 1.0 g remains, so the water formed has a mass of 18.0 g.
B18.0 gCorrect n(O₂) = 16.0/32.00 = 0.500 mol and n(H₂) = 3.00/2.016 = 1.49 mol. O₂ needs twice its moles of H₂, 1.00 mol, which is available, so O₂ is limiting. It forms 2 × 0.500 = 1.00 mol H₂O, or 18.0 g; about 1.0 g of H₂ is left over.
C26.8 g A student who takes the reactant with the smaller mass, H₂, as limiting picks this: 1.49 mol H₂ → 1.49 mol H₂O = 26.8 g. In moles there is more H₂ (1.49 mol) than the 1.00 mol the O₂ can use, so O₂ is limiting.
D16.0 g A student who thinks the mass of a product equals the mass of the reactant it came from picks this, taking the 16.0 g of O₂ that reacts as the mass of water. The water also contains the 2.0 g of H₂ that reacts: 1.00 mol H₂O is 18.0 g.
Working n(H₂) = 3.00 g ÷ 2.016 g/mol = 1.49 mol; n(O₂) = 16.0 g ÷ 32.00 g/mol = 0.500 mol. 0.500 mol O₂ needs 1.00 mol H₂, and 1.49 mol is present, so O₂ is limiting. n(H₂O) = 2 × 0.500 = 1.00 mol; mass = 1.00 mol × 18.02 g/mol = 18.0 g. (H₂ left over: 0.49 mol, 0.98 g.)
The diagram labeled Before reaction represents a mixture of N₂ and H₂ molecules in a sealed container. The gases react according to the equation N₂(g) + 3 H₂(g) → 2 NH₃(g). Which numbered diagram best represents the contents of the container after the reaction has gone to completion?
Answer and reasoning
ADiagram 1Correct Each N₂ molecule reacts with three H₂ molecules, so the 6 H₂ molecules react with 2 N₂ molecules: H₂ is limiting. They form 4 NH₃ molecules, and 1 N₂ molecule is left over. All 6 N atoms and 12 H atoms are accounted for.
BDiagram 2 A student who thinks every reactant is used up when a reaction goes to completion picks this. Only 2 of the 3 N₂ molecules are needed for the 6 H₂ molecules, so one N₂ molecule must remain; this diagram has lost two N atoms.
CDiagram 3 A student who takes the reactant present in fewer molecules, N₂, as the limiting reactant picks this, converting all 3 N₂ into 6 NH₃. That would need 9 H₂ molecules, but only 6 are present, and the diagram shows 18 H atoms where there were 12.
DDiagram 4 A student who reads the coefficient in 2 NH₃ as forming one N₂H₆ molecule picks this. The coefficient 2 means two separate NH₃ molecules, so the product should be drawn as NH₃ molecules, each with one N atom and three H atoms.
Working Before: 3 N₂ and 6 H₂. Each N₂ needs 3 H₂, so 6 H₂ react with 2 N₂: H₂ is limiting. 6 H₂ + 2 N₂ → 4 NH₃, leaving 1 N₂. Check atoms: N 6 = 4 + 2; H 12 = 12. After: 4 NH₃ + 1 N₂.
A 2.70 g sample of Al(s) reacts completely with excess HCl(aq) according to the equation 2 Al(s) + 6 HCl(aq) → 2 AlCl₃(aq) + 3 H₂(g). What volume of H₂(g) is produced, measured at 25°C and 1.00 atm?
Answer and reasoning
A3.36 L A student who uses 22.4 L/mol for a gas under any conditions picks this: 0.150 mol × 22.4 L/mol = 3.36 L. That molar volume applies at 273 K and 1.0 atm; at 298 K the gas occupies more volume, 3.67 L.
B2.45 L A student who takes the moles of product to equal the moles of reactant picks this, using 0.100 mol H₂. The equation gives 3 mol H₂ for every 2 mol Al, so 0.150 mol H₂ forms.
C32.8 L A student who thinks the mass of a product equals the mass of the reactant picks this, taking 2.70 g of H₂ (1.34 mol). Only the moles of H₂ follow from the moles of Al; 0.150 mol H₂ has a mass of only 0.302 g.
D3.67 LCorrect n(Al) = 2.70/26.98 = 0.100 mol, and the equation gives 3 mol H₂ per 2 mol Al, so n(H₂) = 0.150 mol. V = nRT/P = 0.150 × 0.08206 × 298 ÷ 1.00 = 3.67 L.
Working n(Al) = 2.70 g ÷ 26.98 g/mol = 0.100 mol. n(H₂) = 0.100 × 3/2 = 0.150 mol. V = nRT/P = (0.150 mol)(0.08206 L·atm/(mol·K))(298 K) ÷ 1.00 atm = 3.67 L.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
4.5.A.1 Conservation of atoms Fix
Conservation of atoms
In a chemical reaction atoms are rearranged but not created or destroyed, so the number of atoms of each element, and the total mass, are the same before and after the reaction in a closed system. The number of molecules or moles is not necessarily conserved.
Limiting reactant
The reactant that is used up first when a reaction goes to completion; it determines the maximum amount of product. It is found by comparing the moles available with the mole ratio from the balanced equation, not by comparing masses, coefficients or numbers of moles alone.
Excess reactant
A reactant present in more than the amount needed to react with all of the limiting reactant; some of it remains unreacted after the reaction is complete.
Students often think When a reaction goes to completion, all of every reactant is converted into products, so no reactant is left over. In fact Not unless they are mixed in exactly the mole ratio of the balanced equation. The limiting reactant is used up, and whatever is left of the other reactant remains unchanged alongside the products.
Students often think The reactant with the smaller mass is the limiting reactant. In fact Not necessarily. Equal masses of different substances contain different numbers of moles, and the mole ratio also matters. For 2 H₂ + O₂ → 2 H₂O, 4.00 g of H₂ is about 2 mol, more than the 1.00 mol that the 0.500 mol of O₂ in 16.0 g can react with, so O₂ is limiting although its mass is larger.
4.5.A.2 Mole ratio Fix
Mole ratio
The ratio of the coefficients of two substances in a balanced equation; it gives the ratio of the numbers of particles, and of moles, of the substances that react or form. It is not a mass ratio.
Maximum (theoretical) amount of product
The amount of product formed when all of the limiting reactant reacts, calculated from the moles of limiting reactant and the mole ratio; it is converted to mass with the molar mass of the product.
Students often think The reactant present in the smaller number of moles or molecules is the limiting reactant. In fact Not necessarily. Which reactant runs out first depends on the moles present compared with the mole ratio. For N₂ + 3 H₂ → 2 NH₃, 3 mol N₂ with 6 mol H₂ leaves N₂ in excess even though fewer moles of N₂ are present, because each N₂ needs three H₂.
Students often think The reactant with the smaller coefficient in the balanced equation is the limiting reactant, because less of it is needed or present. In fact No. Coefficients give the ratio in which substances react, not the amounts present. Which reactant is limiting depends on the moles actually present compared with that ratio.
4.5.A.3 Gas stoichiometry Fix
Gas stoichiometry
Combining mole ratios with the ideal gas law, PV = nRT, to relate the amount of a gaseous reactant or product to its volume, pressure and temperature (in kelvins). The molar volume 22.4 L/mol applies only at STP (273.15 K and 1.0 atm).
Solution stoichiometry
Combining mole ratios with molarity, M = n/V (V in liters of solution), to relate amounts of dissolved reactants and products; for an ionic solute, the moles of each ion follow from the formula (0.100 M CaCl₂ contains 0.200 M Cl⁻).
Students often think One mole of any gas occupies 22.4 L, whatever its temperature and pressure. In fact No. 22.4 L/mol is the molar volume of an ideal gas at STP (273.15 K and 1.0 atm). At other temperatures or pressures the volume must be found from PV = nRT.
Students often think Dissolving one mole of an ionic compound gives one mole of each kind of ion, whatever the subscripts in its formula. In fact No. The formula gives the number of each ion per formula unit: 0.00250 mol CaCl₂ gives 0.00250 mol Ca²⁺ and 0.00500 mol Cl⁻.
8 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 8
Excess AgNO₃(aq) is added to 25.0 mL of 0.100 M CaCl₂(aq), and all of the chloride ions precipitate as AgCl(s) (molar mass 143.32 g/mol). What mass of AgCl(s) forms?
Answer and reasoning
A0.358 g A student who takes the moles of Cl⁻ to equal the moles of CaCl₂ picks this, using 0.00250 mol AgCl. CaCl₂(s) → Ca²⁺(aq) + 2 Cl⁻(aq), so the solution contains 0.00500 mol Cl⁻.
B0.277 g A student who thinks the mass of product equals the mass of a reactant picks this: 0.00250 mol × 110.98 g/mol = 0.277 g is the mass of CaCl₂ dissolved. The mass of AgCl follows from the moles of Cl⁻ and the molar mass of AgCl.
C0.717 gCorrect n(CaCl₂) = 0.0250 L × 0.100 M = 0.00250 mol. Each formula unit gives two Cl⁻ ions, so n(Cl⁻) = 0.00500 mol, and each Cl⁻ forms one AgCl: 0.00500 mol × 143.32 g/mol = 0.717 g.
D0.555 g A student who reads the coefficients of CaCl₂ + 2 AgNO₃ → 2 AgCl + Ca(NO₃)₂ as a mass ratio picks this, doubling the 0.277 g of CaCl₂. Coefficients give mole ratios: 0.00250 mol CaCl₂ gives 0.00500 mol AgCl, or 0.717 g.
Working n(CaCl₂) = 0.0250 L × 0.100 mol/L = 0.00250 mol, so n(Cl⁻) = 2 × 0.00250 = 0.00500 mol. Ag⁺(aq) + Cl⁻(aq) → AgCl(s): n(AgCl) = 0.00500 mol; mass = 0.00500 × 143.32 = 0.717 g.
Equal masses of methane, CH₄ (molar mass 16.04 g/mol), and ethane, C₂H₆ (molar mass 30.07 g/mol), are burned completely in excess oxygen: CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l) and 2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(l). What is the ratio (moles of CO₂ from the ethane)/(moles of CO₂ from the methane)?
Answer and reasoning
A2.00 A student who thinks equal masses contain equal numbers of moles picks this, counting only the two carbon atoms per ethane molecule. A gram of ethane contains fewer molecules than a gram of methane (16.04/30.07 = 0.533 times as many), so the ratio is 2 × 0.533 = 1.07.
B1.07Correct Moles of CO₂ = (mass/molar mass) × (mol CO₂ per mol fuel). Ethane gives 2 mol CO₂ per mole but has the larger molar mass: ratio = (2/30.07)/(1/16.04) = 2 × 16.04/30.07 = 1.07.
C1.00 A student who thinks equal masses of reactant give equal amounts of product, because mass is conserved, picks this. The mass of CO₂ formed depends on how many carbon atoms each fuel supplies, not just on the mass burned.
D0.53 A student who takes one mole of product per mole of reactant picks this, comparing only the moles of fuel: 16.04/30.07 = 0.53. The equations give 2 mol CO₂ per mole of ethane but 1 mol CO₂ per mole of methane, so the ratio is 1.07.
Working For a mass m: n(CO₂ from CH₄) = (m/16.04) × 1; n(CO₂ from C₂H₆) = (m/30.07) × 4/2 = 2m/30.07. Ratio = (2/30.07)/(1/16.04) = 2 × 16.04/30.07 = 1.07. Each ethane molecule gives twice as much CO₂, but a gram of ethane contains only 16.04/30.07 = 0.533 times as many molecules.
Samples of Mg(s) of different masses were each added to a separate 25.0 mL portion of the same HCl(aq) solution. The reaction is Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g). The graph shows the volume of H₂(g) produced in each trial, measured at 25°C and 1.00 atm. What is the molarity of the HCl(aq)?
Answer and reasoning
A0.100 M A student who takes the moles of HCl to equal the moles of H₂ picks this. The equation shows 2 mol HCl for each mole of H₂, so the 0.00250 mol of H₂ came from 0.00500 mol HCl.
B0.219 M A student who converts the 61.2 mL of H₂ with 22.4 L/mol picks this, getting 0.00273 mol H₂. The gas was measured at 298 K, not at STP, so n = PV/RT = 0.00250 mol.
C0.200 MCorrect The graph levels off at 61.2 mL because the HCl is used up; extra Mg does not react. n(H₂) = PV/RT = (1.00)(0.0612) ÷ (0.08206 × 298) = 0.00250 mol, so n(HCl) = 2 × 0.00250 = 0.00500 mol, and 0.00500 mol ÷ 0.0250 L = 0.200 M.
D0.329 M A student who thinks every reactant is used up completely picks this, assuming all 0.100 g of Mg in the last trial reacted (0.00412 mol, needing 0.00823 mol HCl). The flat part of the graph shows that the extra Mg did not react because the HCl had run out.
Working The volume levels off at 61.2 mL: beyond this point HCl is limiting and extra Mg does not react. n(H₂) = PV/RT = (1.00 atm)(0.0612 L) ÷ [(0.08206 L·atm/(mol·K))(298 K)] = 0.00250 mol. n(HCl) = 2 × 0.00250 = 0.00500 mol; M = 0.00500 mol ÷ 0.0250 L = 0.200 M.
A student wants to determine the mass of CO₂(g) produced when a weighed sample of CaCO₃(s) reacts completely with excess HCl(aq): CaCO₃(s) + 2 HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). Which procedure would give the mass of CO₂ produced?
Answer and reasoning
AWeigh the open flask and its contents before and after the reaction and find the loss in massCorrect All the atoms are conserved, and the only product that leaves an open flask is CO₂ gas, so the decrease in mass of the flask and its contents equals the mass of CO₂ produced.
BWeigh the tightly stoppered flask and its contents before and after the reaction A student who thinks a gas produced in a closed container makes the mass decrease picks this. In a stoppered flask the CO₂ stays inside and has mass, so the total mass does not change and gives no measure of the CO₂.
CMeasure the volume of the excess HCl(aq) added, and calculate the mass of CO₂ from that volume A student who thinks the amount of product depends on the reactant in excess picks this. The HCl is in excess, so the CO₂ formed is set by the CaCO₃, not by how much HCl was added.
DWeigh the sample of CaCO₃ before it reacts, and take the mass of CO₂ produced to equal that mass A student who thinks a product has the same mass as the reactant it came from picks this. Only some of the atoms of CaCO₃ end up in CO₂: 100.09 g of CaCO₃ gives 44.01 g of CO₂.
Working No calculation. Mass is conserved; in an open flask the only matter that leaves is the CO₂ gas, so the decrease in mass of the flask and contents equals the mass of CO₂ released. In a stoppered flask the mass does not change.
Carbon monoxide reacts with oxygen according to the equation 2 CO(g) + O₂(g) → 2 CO₂(g). A student is asked to draw the contents of the container after the mixture shown in the Before diagram reacts completely. Which statement correctly evaluates the student's diagram?
Answer and reasoning
AIt is consistent, because the number of molecules is the same as in the Before diagram A student who thinks the number of molecules is conserved in a reaction picks this. Atoms, not molecules, are conserved: the correct after diagram has 5 molecules (4 CO₂ and 1 O₂), and the student's diagram has 4 more O atoms than the mixture started with.
BIt is not consistent, because all of the O₂ should have been used up in the reaction A student who thinks every reactant is used up when a reaction goes to completion picks this. The 4 CO molecules need only 2 of the 3 O₂ molecules, so one O₂ molecule should remain.
CIt is not consistent, because there are fewer O₂ molecules, so some CO should be left A student who takes the reactant with fewer molecules as limiting picks this. Each O₂ reacts with two CO, so 3 O₂ could react with 6 CO; only 4 CO are present, so CO is limiting and none of it remains.
DIt is not consistent, because it contains more O atoms than the Before diagram doesCorrect Before the reaction there are 4 C atoms and 10 O atoms. The 4 CO molecules use 2 O₂ to form 4 CO₂, leaving 1 O₂: 10 O atoms in all. The student's diagram shows 4 CO₂ and 3 O₂, which is 14 O atoms, so atoms are not conserved.
Working Before: 4 CO + 3 O₂ = 4 C atoms and 4 + 6 = 10 O atoms. 4 CO need 2 O₂, so CO is limiting: products 4 CO₂ with 1 O₂ left (4 C, 8 + 2 = 10 O). The student's diagram, 4 CO₂ + 3 O₂, contains 8 + 6 = 14 O atoms: atoms are not conserved.
A 12.0 g sample of Mg(s) and 12.0 g of O₂(g) are allowed to react according to the equation 2 Mg(s) + O₂(g) → 2 MgO(s). Which statement about the limiting reactant is correct and correctly justified?
Answer and reasoning
AMg is limiting, because its 0.494 mol needs only 0.247 mol of O₂, and 0.375 mol of O₂ is presentCorrect n(Mg) = 12.0/24.30 = 0.494 mol and n(O₂) = 12.0/32.00 = 0.375 mol. The mole ratio is 2 Mg : 1 O₂, so 0.494 mol Mg needs 0.247 mol O₂. More O₂ than that is present, so the Mg runs out first.
BO₂ is limiting, because fewer moles of O₂ (0.375 mol) than of Mg (0.494 mol) are present A student who takes the reactant present in fewer moles as limiting picks this. The comparison must use the mole ratio: each O₂ reacts with two Mg, so 0.375 mol O₂ could react with 0.750 mol Mg, more than is present.
CO₂ is limiting, because its coefficient (1) is smaller than the coefficient of Mg (2) A student who takes the reactant with the smaller coefficient as limiting picks this. Coefficients give the ratio in which the substances react, not the amounts present; comparing the 0.375 mol O₂ with the 0.247 mol needed shows O₂ is in excess.
DNeither is limiting, because both reactants are completely converted into MgO A student who thinks all reactants are used up when a reaction goes to completion picks this. The amounts are not in the 2 : 1 mole ratio: Mg runs out when only 0.247 mol of the 0.375 mol O₂ has reacted.
Working n(Mg) = 12.0 ÷ 24.30 = 0.494 mol; n(O₂) = 12.0 ÷ 32.00 = 0.375 mol. 0.494 mol Mg needs 0.494/2 = 0.247 mol O₂, less than the 0.375 mol present, so Mg is limiting and 0.128 mol O₂ remains.
A rigid, sealed container holds 0.300 mol of H₂(g) and 0.200 mol of O₂(g) at 150°C, and the total pressure is 2.00 atm. A spark is used to start the reaction 2 H₂(g) + O₂(g) → 2 H₂O(g), which goes to completion. The container is then returned to 150°C, a temperature at which all of the water remains a gas. What is the total pressure in the container after the reaction?
Answer and reasoning
A1.33 atm A student who thinks every reactant is used up completely picks this, treating all 0.500 mol of the mixture as converted in the 3 : 2 ratio of gas moles in the equation: 2.00 atm × 2/3 = 1.33 atm. The 0.300 mol of H₂ uses only 0.150 mol of the O₂, so 0.050 mol of O₂ remains and 0.350 mol of gas is present.
B1.40 atmCorrect H₂ is limiting: 0.300 mol H₂ reacts with 0.150 mol O₂ to form 0.300 mol H₂O(g), and 0.050 mol O₂ is left over. The moles of gas fall from 0.500 mol to 0.350 mol, and at constant volume and temperature the pressure is proportional to the moles of gas: 2.00 atm × 0.350/0.500 = 1.40 atm.
C1.60 atm A student who takes the reactant present in fewer moles, O₂, as the limiting reactant picks this: 0.200 mol O₂ would form 0.400 mol H₂O, giving 2.00 atm × 0.400/0.500 = 1.60 atm. That would need 0.400 mol H₂, and only 0.300 mol is present, so H₂ is limiting.
D2.00 atm A student who thinks the number of moles of gas is conserved in a reaction picks this, leaving the pressure unchanged. Atoms are conserved, but three moles of reactant gas form two moles of H₂O(g), so the total moles of gas, and the pressure, decrease.
Working 0.300 mol H₂ needs 0.150 mol O₂, and 0.200 mol is present, so H₂ is limiting. After the reaction: 0.300 mol H₂O(g) and 0.200 − 0.150 = 0.050 mol O₂, a total of 0.350 mol of gas. Before the reaction the total was 0.500 mol. At constant volume and temperature, P is proportional to n (PV = nRT): P = 2.00 atm × (0.350 mol ÷ 0.500 mol) = 1.40 atm. Distractors: all 0.500 mol of the mixture taken to react in the 3 : 2 ratio of gas moles, 2.00 × 2/3 = 1.33 atm; O₂ taken as limiting, 0.400 mol H₂O, 2.00 × 0.400/0.500 = 1.60 atm; moles of gas taken as unchanged, 2.00 atm.
A student needs to prepare 6.76 g of solid silver phosphate, Ag₃PO₄ (molar mass 418.58 g/mol), by adding AgNO₃(aq) to excess Na₃PO₄(aq): 3 AgNO₃(aq) + Na₃PO₄(aq) → Ag₃PO₄(s) + 3 NaNO₃(aq). What is the minimum volume of 0.150 M AgNO₃(aq) (molar mass of AgNO₃ 169.88 g/mol) that is needed?
Answer and reasoning
A323 mLCorrect n(Ag₃PO₄) = 6.76 g ÷ 418.58 g/mol = 0.01615 mol. Each formula unit contains three Ag atoms, so the equation requires 3 mol AgNO₃ per mole of Ag₃PO₄: 0.04845 mol. V = 0.04845 mol ÷ 0.150 M = 0.323 L = 323 mL.
B108 mL A student who relates reactant and product one to one picks this, using 0.01615 mol AgNO₃: 0.01615 mol ÷ 0.150 M = 108 mL. The equation shows 3 mol AgNO₃ for each mole of Ag₃PO₄, so three times this volume is needed.
C265 mL A student who thinks the mass of a product equals the mass of the reactant used picks this, taking 6.76 g of AgNO₃: 6.76 g ÷ 169.88 g/mol = 0.0398 mol, or 265 mL. The amount of AgNO₃ follows from the moles of Ag₃PO₄ and the 3 : 1 mole ratio, not from equal masses.
D796 mL A student who reads the coefficients as a mass ratio picks this, taking 3 × 6.76 g = 20.28 g of AgNO₃, which is 0.119 mol, or 796 mL. The coefficients give the ratio of moles: 0.01615 mol Ag₃PO₄ needs 0.04845 mol AgNO₃.
Working n(Ag₃PO₄) = 6.76 g ÷ 418.58 g/mol = 0.01615 mol. n(AgNO₃) = 3 × 0.01615 = 0.04845 mol. V = 0.04845 mol ÷ 0.150 mol/L = 0.323 L = 323 mL. Distractors: 1 : 1 mole ratio, 0.01615 mol ÷ 0.150 M = 108 mL; mass of AgNO₃ taken as 6.76 g, 6.76/169.88 = 0.03979 mol, 265 mL; mass of AgNO₃ taken as 3 × 6.76 g = 20.28 g, 0.1194 mol, 796 mL.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
