Stoichiometry and Solutions Practice

Introduction

Stoichiometry is a chain: identify species, balance the net reaction, convert measurements to moles, compare reactants by coefficients and convert the result to the requested quantity. Solution problems add a concentration denominator and sometimes a dilution step. Many wrong answers are numerically plausible because one link is skipped. This practice page emphasizes the chain and requires a check at every transition.

Core explanation

In a balanced equation, coefficients represent ratios of amounts in moles for the reaction as written. Subscripts belong to chemical formulas and must not be changed to achieve balance. For 2Al + 3Cl₂ → 2AlCl₃, three moles Cl₂ are needed per two moles Al. If both reactants have finite amounts, compare n/ν for each, where ν is its reactant coefficient; the smaller possible reaction extent sets the limiting reagent . The theoretical product amount uses the product coefficient times that extent. Actual yield can be lower because of incomplete conversion, side reactions, separation loss or impurity.

For a solution, molar concentration c = n/V solution uses final solution volume. Multiplying concentration by volume gives solute amount when units match: 0.100 mol L⁻¹ × 0.250 L = 0.0250 mol. If solvent is added without loss or reaction of solute, n initial = n final , leading to c₁V₁ = c₂V₂ for compatible volume units. The equality is not a universal mixing law: volumes may be nonadditive, solute may react, and concentration might be defined on a different basis. A dilution changes concentration, not solute amount. A reaction between two solutions requires stoichiometry after calculating their solute amounts.

To prepare a solution by mass, first calculate target amount n = cV , then required mass m = nM . Dissolve and make up to the final calibration mark; adding the calculated solvent volume directly to solid can produce the wrong final solution volume. If the reagent has purity less than 100%, the mass weighed must be adjusted by the assayed mass fraction, with uncertainty and safety procedures accounted for.

Step-by-step reasoning

1. Balance the equation with correct formulas and states. 2. Convert every supplied mass or solution quantity to moles of its named species. 3. Compare each reactant mole amount divided by its coefficient. 4. Multiply the limiting reaction extent by the product coefficient. 5. Convert to requested mass, concentration or percent yield with units. 6. For a dilution, verify that no solute is added, lost or reacted before using c₁V₁ = c₂V₂ .

Visual explanation

Picture two funnels labeled reactant A and B pouring mole amounts into a reaction box. Each funnel has a gate marked by its coefficient; the funnel supporting fewer complete reaction units limits output. A separate flask diagram shows a fixed number of solute dots before and after adding solvent: the final flask is larger but contains the same solute dots, explaining dilution.

Real-world analogy

Assembling kits from two screws and one bracket per kit illustrates the limiting component. Adding water to a drink spreads the same sugar over a larger volume and illustrates dilution. Neither analogy captures side reactions or solution-volume nonadditivity, so the chemical conditions must still be checked.

Real-world example

A pharmacy prepares a standard solution by transferring a measured aliquot of concentrated stock to a volumetric flask and filling to the mark. The operator records stock concentration, aliquot volume and final calibrated volume. If the solute is stable and no material is lost, amount conservation gives the target concentration. Calling the added water volume the final volume would be a procedural error; the flask mark defines the final solution volume.

Why?

Why compare normalized moles rather than grams? Chemical equations count entities, and different substances have different molar masses. Even comparing raw moles can mislead if coefficients differ. Dividing each amount by its coefficient gives a common unit of reaction extent, making the limiting comparison meaningful.

Common misconception

“The smaller mass is the limiting reagent.” Molar masses and coefficients matter. “Dilution removes solute.” It generally changes volume while conserving solute amount. “Theoretical yield is guaranteed.” It assumes complete target conversion. “Molarity uses solvent volume.” It uses final solution volume. “A stock solution formula works even when solute reacts on mixing.” Reaction requires a new mass balance.

Worked example

React 5.40 g Al with 10.0 g Cl₂ in 2Al + 3Cl₂ → 2AlCl₃. Use 26.98 g mol⁻¹ for Al and 70.90 g mol⁻¹ for Cl₂. Amounts are 0.200 mol Al and 0.141 mol Cl₂. Normalized extents are 0.200/2 = 0.100 mol and 0.141/3 = 0.0470 mol, so Cl₂ limits. Theoretical AlCl₃ is 2(0.0470) = 0.0940 mol, about 12.5 g using 133.34 g mol⁻¹. If the product is dissolved to 0.500 L final volume with no loss or secondary chemistry, its formal concentration from that amount is 0.188 mol L⁻¹. A real aqueous AlCl₃ solution has speciation and hydrolysis complexities, so “formal concentration” is the cautious term for total dissolved AlCl₃-derived amount.

Quick check

1. Does doubling final solution volume by adding solvent halve concentration if solute amount is unchanged? Answer: Yes, under the stated dilution assumptions. 2. In 2A + B → C, do 2 mol A and 2 mol B provide equal reaction extents? Answer: No. A supports 1 mol extent, B supports 2 mol; A is limiting.

Exam focus

Write the balanced equation and each molar mass. Show mass-to-mole or concentration-to-mole conversions, then normalized extents. Distinguish theoretical from actual yield. Use final solution volume and specify whether the calculated concentration is formal total or a particular equilibrium species.

Advanced insight

In solution, volume can change upon mixing, and reactive solutes may have several species. Analytical concentration tracks total amount of an element or input formula, whereas equilibrium concentration of a free ion may be much lower. A rigorous material balance includes all species sharing that component and may require equilibrium calculations after the initial stoichiometry.

Summary

Balanced coefficients govern amount ratios; normalized mole amounts identify the limiting reagent. Concentration relates solute amount to final solution volume, and dilution conserves amount under stated assumptions. A clear chain of units and definitions prevents hidden errors.

Practice questions

1. In N₂ + 3H₂ → 2NH₃, what maximum NH₃ amount comes from 0.50 mol N₂ with excess H₂? Answer: 1.0 mol NH₃. 2. What volume of 0.200 mol L⁻¹ solution contains 0.0500 mol solute? Answer: V = 0.0500/0.200 = 0.250 L. 3. Dilute 25.0 mL of 2.00 mol L⁻¹ stock to 100.0 mL. Find final concentration. Answer: (2.00 × 25.0)/100.0 = 0.500 mol L⁻¹. 4. Why is 5.00 g of reagent not necessarily less chemical amount than 10.0 g of another? Answer: Their molar masses differ; convert each mass to moles before comparing.