# Chapter 101, module 1 supplement: Stoichiometry bridge

Original prerequisite practice for quantitative science. This is not an official exam question set or a complete chemistry lesson. Planning-ledger link: C02. The supplement applies the unit and ratio reasoning of module 101-1; it does not change that narration module or claim that its existing cards cover chemistry.

## Reasoning framework

An amount in moles counts chemical entities; mass measures a different quantity. Convert mass to amount by dividing by molar mass, keeping units visible. A balanced reaction supplies ratios of amounts, not generally ratios of masses. Determine how much product each reactant could support, then choose the smaller amount. An excess reactant cannot compensate for exhaustion of another required reactant.

Distinguish theoretical yield from measured yield. The theoretical yield uses the balanced equation and available reactants under stated assumptions. A measured mass may include impurities or retained solvent. A calculation above one hundred percent is a warning to inspect the assumptions and measurement, not proof that conservation of matter failed. These exercises specify a simplified reaction so that the reasoning is self-contained.

## Passage

In an idealised reaction, A plus two B produces one C. The molar masses are 40 grams per mole for A, 30 grams per mole for B and 100 grams per mole for C. There are no other products. A technician combines 8.0 grams of pure A with 9.0 grams of pure B. Assume complete reaction of the limiting reactant, no side reactions and no material lost during reaction. Unreacted starting material is removed before the product is weighed. The dry isolated product weighs 12 grams.

In a second run, the technician uses the same 8.0 grams of A but twice as much B. In a third run, the original quantities are used, but the isolated product is weighed before drying and reads 16 grams.

## Questions

1. Convert the original reactant masses to moles. Explain why comparing 8.0 grams directly with 9.0 grams does not identify the limiting reactant.
2. Which reactant limits the original reaction? Calculate the maximum amount and mass of C.
3. Calculate the amount and mass of A left over. Check the total mass balance before product isolation.
4. What is the isolated percentage yield of the dry product? Does a lower isolated yield alone prove incomplete chemical conversion?
5. In the second run, what is the theoretical mass of C? Why does doubling B not double the original theoretical yield?
6. Calculate the apparent percentage yield in the third run. Give an explanation consistent with conservation of matter.
7. Dissolve the 12 grams of pure dry C to a final volume of 0.60 litres. What is the molar concentration? State which assumption about the isolated solid this requires.
8. Transfer 25 millilitres of that solution and dilute to a final volume of 100 millilitres. Calculate the new concentration and the amount of C transferred.

## Worked answers

1. A: 8.0 divided by 40 equals 0.20 moles. B: 9.0 divided by 30 equals 0.30 moles. Different molar masses and reaction coefficients make raw mass comparison inappropriate.

2. The available A could support 0.20 moles of C. The available B supports only 0.30 divided by two, or 0.15 moles of C. B is limiting. Theoretical product mass is 0.15 times 100, or 15 grams.

3. Formation of 0.15 moles of C consumes 0.15 moles of A. Remaining A is 0.05 moles, or 2.0 grams. Initially there are 17 grams of reactants. After reaction there are 15 grams of C and 2 grams of A. Removing the excess A during isolation does not destroy its mass; it leaves the product fraction.

4. Twelve divided by fifteen times one hundred gives 80 percent. Product lost during transfer or purification could reduce isolated yield even if the limiting reactant reacted completely. Yield alone does not identify the cause.

5. Eighteen grams of B is 0.60 moles, enough for 0.30 moles of C. A now limits production to 0.20 moles, or 20 grams. The limiting reactant changes; the result is not 30 grams.

6. Sixteen divided by fifteen times one hundred is approximately 106.7 percent. Retained solvent adds to the measured mass but is not C. The measurement is not a valid mass of pure dry product.

7. Twelve grams divided by 100 grams per mole is 0.12 moles. Divide by the final 0.60 litres to obtain 0.20 moles per litre. This requires the weighed material to be pure C; impurities would invalidate that amount calculation.

8. The concentration is 0.20 times 25 divided by 100, or 0.050 moles per litre. Transferred amount is 0.20 times 0.025 litres, or 0.0050 moles, equal to 5.0 millimoles. Dilution preserves that amount under the no-loss assumption.

## Review boundary

Arithmetic is checked separately from editorial review. Independent chemistry review, Japanese semantic review, learner difficulty calibration and exam-topic crosswalk remain pending. No new recall cards are implied by this supplement. Follow-up lesson requirements: chemical entities, molar mass, equation balancing, limiting reagents, purity, yields and concentration conversions.