AP Chem 4.5: Stoichiometry

Every conversion goes through moles

Chapter 2.6 used mole ratios for calculations. AP stoichiometry questions mix three ways of giving an amount:

n = \frac{m}{M} \qquad n = MV \qquad n = \frac{PV}{RT}

Whether you’re given a mass, a solution’s molarity and volume, or a gas volume, convert to moles first, use the coefficient ratio, and convert back to whatever the question asks for.

The limiting reactant: moles ÷ coefficient

When amounts of both reactants are given, the one that runs out first is the limiting reactant限量试剂; it determines how much product forms. The other is the excess reactant过量试剂, and some of it is left over. To decide:

divide each reactant’s moles by its coefficient; the smallest result marks the limiting reactant;
comparing masses alone, or moles alone, doesn’t work; the coefficients have to be included.

The before–change–after table

AP teachers often organize the calculation in a three-row table, called a BCA table:

	Reactant 1	Reactant 2	Product
Before (mol)	given	given	0
Change (mol)	−coefficient × x	−coefficient × x	+coefficient × x
After (mol)	before − change	before − change	coefficient × x

Here x is the limiting reactant’s moles ÷ coefficient. The change row always follows the coefficient ratio, and the limiting reactant ends at 0.

Choose a reaction and change the amounts of the reactants to see how the table and results change:

Try it: stoichiometry with a before–change–after table

Reaction:

\ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)}

Gas temperature (°C)Gas pressure (atm)

Particle diagrams of a limiting reactant

A particle diagram shows directly which reactant runs out: combine particles in the coefficient ratio as many times as possible, and what’s left is the excess reactant.

Try it: particle view — which runs out first?

Before

After

HOC

Percent yield

The amount of product calculated from the limiting reactant is the theoretical yield; the amount actually obtained is the actual yield (2.6.7):

\text{percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%

A percent yield above 100% usually means the product wasn’t dried (it still contains water) or contains impurities.

Gas reactions: volume and pressure

At the same temperature and pressure, gas volumes are in the same ratio as moles, so volumes can be used directly (Example 2). At constant temperature in a rigid container, the pressure of the gas is proportional to its moles: for example, when SO₂ and O₂ react exactly in a 2 : 1 ratio to form SO₃, 3 parts of gas become 2, so the total pressure drops to 2/3 of its starting value.

Example 1: A metal and an acid forming a gas

1.31 g of zinc is added to 50.0 mL of 0.500 M hydrochloric acid: $\ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)}$ . Find the volume of H₂ formed at 25 °C and 1.00 atm, and the amount of zinc left over (in moles).

Step-by-step solution

Convert to moles. Coefficients are mole ratios, so turn both the mass and the solution volume into moles:
$\begin{aligned} n(\text{Zn}) &= \frac{1.31}{65.38} = 0.0200 \\ n(\text{HCl}) &= 0.500 \times 0.0500 \\ &= 0.0250 \end{aligned}$
Find the limiting reactant. Divide each reactant’s moles by its coefficient; the smaller result runs out first:
$\begin{aligned} \text{Zn} &: 0.0200 \div 1 = 0.0200 \\ \text{HCl} &: 0.0250 \div 2 = 0.0125 \end{aligned}$
HCl gives the smaller result, so it is the limiting reactant and Zn is in excess.
Before–change–after table. All the HCl reacts, and Zn and $\ce{H2}$ each change by half as much (in mmol; $\ce{ZnCl2}$ matches $\ce{H2}$ and is left out):

Zn HCl H₂
Before 20.0 25.0 0
Change −12.5 −25.0 +12.5
After 7.5 0 12.5

The zinc left over is $\boxed{7.5 \times 10^{-3}\ \text{mol}}$ .
Volume of H₂. Use V = nRT/P with T = 298.15 K and R = 0.08206 L·atm/(mol·K):
$\begin{aligned} V &= \frac{nRT}{P} \\ &= \frac{(0.0125)(0.08206)(298.15)}{1.00} \\ &= \boxed{0.306\ \text{L}} \end{aligned}$
Check. Try Zn as the limiting reactant: 0.0200 mol Zn would need 0.0400 mol HCl, and only 0.0250 mol is present, so HCl really is limiting.

	Zn	HCl	H₂
Before	20.0	25.0	0
Change	−12.5	−25.0	+12.5
After	7.5	0	12.5

Example 2: Gases at the same temperature and pressure

At the same temperature and pressure, 2.00 L of CH₄ and 3.00 L of O₂ are mixed and ignited: $\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$ . Which gases remain afterward, and what are their volumes at the same temperature and pressure?

Step-by-step solution

Use volumes in place of moles. At the same temperature and pressure, gas volumes are in the same ratio as moles (Avogadro’s law), so we can work directly in liters.
Find the limiting reactant. Divide each volume by its coefficient:
$\begin{aligned} \ce{CH4} &: 2.00 \div 1 = 2.00 \\ \ce{O2} &: 3.00 \div 2 = 1.50 \end{aligned}$
$\ce{O2}$ gives the smaller result, so it is the limiting reactant.
Before–change–after table. All 3.00 L of $\ce{O2}$ reacts, using 1.50 L of $\ce{CH4}$ and forming 1.50 L of $\ce{CO2}$ (the water is liquid and isn’t counted as gas; volumes in L):

CH₄ O₂ CO₂
Before 2.00 3.00 0
Change −1.50 −3.00 +1.50
After 0.50 0 1.50
State the result. $\boxed{0.50\ \text{L CH}_4}$ remains along with $\boxed{1.50\ \text{L CO}_2}$ , 2.00 L of gas in all.
Check carbon. Before: carbon in 2.00 L of CH₄. After: 0.50 L of CH₄ + 1.50 L of CO₂ = 2.00 L, so carbon is conserved.

	CH₄	O₂	CO₂
Before	2.00	3.00	0
Change	−1.50	−3.00	+1.50
After	0.50	0	1.50

Practice: which reactant is limiting?

2H₂ + O₂ → 2H₂O: 3.0 mol H₂, 1.0 mol O₂
N₂ + 3H₂ → 2NH₃: 2.0 mol N₂, 3.0 mol H₂
N₂ + 3H₂ → 2NH₃: 1.0 mol N₂, 6.0 mol H₂
CH₄ + 2O₂ → CO₂ + 2H₂O: 1.0 mol CH₄, 2.0 mol O₂
Zn + 2HCl → ZnCl₂ + H₂: 0.10 mol Zn, 0.10 mol HCl
2Na + Cl₂ → 2NaCl: 0.40 mol Na, 0.30 mol Cl₂
AgNO₃ + NaCl → AgCl + NaNO₃: 0.0025 mol AgNO₃, 0.0030 mol NaCl
2CO + O₂ → 2CO₂: 4.0 mol CO, 2.0 mol O₂
2H₂ + O₂ → 2H₂O: 4.0 g H₂, 16 g O₂
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂: 0.20 mol CaCO₃, 0.30 mol HCl

Common mistakes

Mistake 1: “The reactant with the smaller mass (or fewer moles) is limiting.” Compare moles ÷ coefficient. In 2H₂ + O₂ → 2H₂O with 4.0 g of H₂ and 16 g of O₂, O₂ has the larger mass but is the limiting reactant.

Mistake 2: “The amount of product can be calculated from either reactant.” Only the limiting reactant gives the right answer. Calculating from the excess reactant overestimates the product.

Mistake 3: “Excess reactant left over = its starting amount − the amount of limiting reactant.” Subtract the amount of the excess reactant actually used, its coefficient × x, not the amount of the limiting reactant.

Quick check

N₂ + 3H₂ → 2NH₃. What is the most NH₃ that 2.0 mol N₂ and 3.0 mol H₂ can form?

2H₂ + O₂ → 2H₂O. What is the greatest mass of water that 4.00 g H₂ and 16.0 g O₂ can form, and why?

At the same temperature and pressure, 5.00 L of H₂ and 5.00 L of Cl₂ react completely: H₂ + Cl₂ → 2HCl. What is the total volume of gas afterward (same temperature and pressure)?

25.0 mL of 0.100 M AgNO₃ is mixed with 25.0 mL of 0.200 M NaCl. Which statement is correct?

When can gas volumes be used in place of moles to find the limiting reactant?

A rigid container holds 3 N₂ molecules and 6 H₂ molecules, and the reaction N₂ + 3H₂ → 2NH₃ goes as far as possible. Which diagram best represents the particles in the container afterward?

2K + Cl₂ → 2KCl. 10.0 g of K reacts with 10.0 g of Cl₂. Which is the limiting reactant, and why?

C₂H₆S burns completely in O₂ to form CO₂, H₂O, and SO₂. When the equation is balanced with the lowest whole-number coefficients, what is the coefficient of O₂?

In a rigid flask, SO₂ and O₂ are mixed in a 2 : 1 ratio at a total pressure of 3.0 atm. They react completely by 2SO₂(g) + O₂(g) → 2SO₃(g), and the flask returns to its starting temperature. What is the final total pressure?

50.0 mL of 0.200 M BaCl₂ is mixed with 50.0 mL of 0.100 M Na₂CO₃, and BaCO₃ precipitates. What is the concentration of Ba²⁺ remaining in solution? (Assume the volumes are additive.)

In one run of Zn + 2HCl → ZnCl₂ + H₂, HCl is the limiting reactant. The run is repeated with twice the mass of Zn and everything else the same. Which statement must be true?

In the same reaction, Zn is always in excess. How should the hydrochloric acid be changed to double the amount of H₂ formed?

1.0 g of each fuel below burns completely. Which one produces the most CO₂?

2.50 g of CuSO₄·xH₂O is heated to constant mass, leaving 1.60 g of anhydrous CuSO₄. What is x?

A reaction has a theoretical yield of 2.50 g, and 2.15 g of product, dried to constant mass, is obtained. What is the percent yield?

AP-style free response

A student mixes 25.0 mL of 0.100 M $\ce{AgNO3}$ with 30.0 mL of 0.100 M NaCl, and a white precipitate forms.

(a) Write the net ionic equation.

(b) Use a before–change–after table to identify the limiting reactant, and calculate the mass of AgCl formed.

(c) Calculate the concentrations of Na⁺, NO₃⁻, and Cl⁻ in the mixture (assume the volumes are additive).

(d) The student filters and rinses the precipitate but weighs it before it is completely dry. Is the measured mass too high or too low? Is the calculated number of moles of AgCl formed too high or too low?

Model answer and scoring

(a) $\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}$

(b) n(Ag⁺) = 0.100 × 0.0250 = 0.00250 mol; n(Cl⁻) = 0.100 × 0.0300 = 0.00300 mol. Both coefficients are 1, so Ag⁺ is the limiting reactant.

	Ag⁺	Cl⁻	AgCl
Before (mol)	0.00250	0.00300	0
Change (mol)	−0.00250	−0.00250	+0.00250
After (mol)	0	0.00050	0.00250

m(AgCl) = 0.00250 mol × 143.32 g/mol = 0.358 g.

(c) Total volume = 55.0 mL = 0.0550 L. [Na⁺] = 0.00300 ÷ 0.0550 = 0.0545 M; [NO₃⁻] = 0.00250 ÷ 0.0550 = 0.0455 M; [Cl⁻] = 0.00050 ÷ 0.0550 = 0.0091 M.

(d) The precipitate still contains water, so the measured mass is too high, and the calculated moles of AgCl are too high as well.

Scoring: (a) net ionic equation; (b) Ag⁺ limiting, correct table, 0.358 g; (c) all three concentrations, using a total volume of 55.0 mL; (d) mass too high and moles too high, with the reason.

Summary

Convert masses, solutions, and gases to moles first: n = m/M, n = MV, n = PV/RT.
The smallest moles ÷ coefficient marks the limiting reactant; product amounts must be calculated from it.
In a before–change–after table, the change row follows the coefficient ratio, and the limiting reactant ends at 0.
For gases at the same temperature and pressure, volume ratios can be used directly.

AP tip: Stoichiometry points are often awarded step by step: converting to moles, using the mole ratio, identifying the limiting reactant, and converting to the requested quantity. Show every step, so a slip at the end doesn’t cost the earlier points. Report the answer to the fewest significant figures in the data.

Stoichiometry化学计量

Learning goals

Before you start

Every conversion goes through moles

The limiting reactant: moles ÷ coefficient

The before–change–after table

Particle diagrams of a limiting reactant

Percent yield

Gas reactions: volume and pressure

Example 1: A metal and an acid forming a gas

Example 2: Gases at the same temperature and pressure

Common mistakes

Quick check

AP-style free response

Summary

Feedback / Report a problem