SCIENCE · INVESTIGATE AND EXPLAIN

Predict product mass without rounding away reactants

Convert masses to moles, apply the reaction ratio, and retain excess material.

Learning outcome: Identify a limiting reactant from stoichiometric amounts and reconcile product plus leftover mass.

Starting recap

For 2 H2 + O2 → 2 H2O, divide H2 moles by two and O2 moles by one to compare supported reaction extent.

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Try

5 g H2 and 32 g O2 correspond to 2.5 mol H2 and 1 mol O2 with the rounded teaching molar masses.

Open the experiment

Predict

Predict water mass and excess hydrogen. Would a smaller mass of hydrogen necessarily make it the limiting reactant?

Your prediction and reason

Experiment

Capture (5,32), (5,64), and (1,32). Check both the total mass ledger and the fractional molar extent. Try zero hydrogen.

Account for continuous molar amounts

Worked teaching model · synthetic values · Ideal complete 2 H2 + O2 → 2 H2O; rounded molar masses 2, 32 and 18 g/mol. Closed accounting boundary, pure reactants and no competing reactions. Molar amounts are continuous; do not round reaction extent down to a whole number.

Initial hydrogen mass: 5 g Initial oxygen mass: 32 g

Reset keeps your prediction, captures and practice history.

Read left to right: each conversion uses the quantity and units from the preceding box. Reaction extent is a continuous amount in moles, not a rounded number of particle groups.

Calculated values on the same fixed scale
Row	Quantity	Value (g)
1	Initial total mass	37
2	Final total mass including leftovers	37
3	Water product mass	36
4	Hydrogen left	1
5	Oxygen left	0

36.000 g water; 1.000 g H2 and 0.000 g O2 remain.

H2 amount = 2.5 mol; O2 amount = 1 mol. Extent = min(n(H2)/2, n(O2)) = 1 mol; limiting inventory: oxygen. This gives 2 mol water. Zero extent means no product; a tie at zero means both inventories are absent, not that a reaction occurred.

Observation to attach to a capture

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Understand

Coefficients compare amounts, not grams

The equation 2 H2 + O2 → 2 H2O specifies a 2:1:2 ratio of molecular amounts. It does not say that two grams of hydrogen react with one gram of oxygen. Convert masses to moles before using the coefficients. Background: reaction stoichiometry, OpenStax Chemistry 2e.

Use the stated rounded molar masses throughout this model: H2 = 2 g/mol, O2 = 32 g/mol and H2O = 18 g/mol. These consistent values preserve the mass ledger: two moles of H2 weigh 4 g, one mole of O2 weighs 32 g, and two moles of water weigh 36 g. Real calculations should use suitable molar masses and precision for the sample and purpose.

Work from a measured starting inventory

With 5 g H2 and 32 g O2, divide each mass by its own molar mass. Hydrogen amount is 5/2 = 2.5 mol; oxygen amount is 32/32 = 1 mol. Compare how much reaction each can support: 2.5/2 = 1.25 mol of extent from hydrogen, versus 1/1 = 1 mol from oxygen. Oxygen limits the ideal extent to 1 mol.

That extent consumes 2 mol H2 and 1 mol O2, producing 2 mol H2O. Product mass is 2 × 18 = 36 g. Hydrogen left is 0.5 mol, or 1 g. Initial mass 37 g equals final water 36 g plus leftover hydrogen 1 g. Looking only at product mass would omit material still inside the boundary.

Let nH and nO be initial molar amounts of H2 and O2. Extent x = min(nH/2, nO). Product amount is 2x; leftover amounts are nH − 2x and nO − x. Multiply each final amount by its own molar mass to calculate a mass ledger.

Fractional moles are ordinary quantities

For 1 g H2 and excess O2, nH = 0.5 mol and x = 0.25 mol. Product amount is 0.5 mol, corresponding to 9 g water. A quarter mole of extent represents an enormous number of reaction events. It is not a quarter of one molecule.

The earlier small-particle lab deliberately used complete groups of a few molecules and rounded group counts down. Do not apply that floor operation to molar amounts. Doing so here would predict zero product from a substantial sample. This is a case where carrying a computational rule outside its model domain gives a wrong scientific conclusion.

Transfer: predict the outcome for 5 g H2 and 64 g O2. Identify the limiting reagent and account for every gram before opening the feedback.

Check the excess-oxygen case

Hydrogen provides 2.5 mol and oxygen provides 2 mol. The extent is min(1.25, 2) = 1.25 mol. Water amount is 2.5 mol, or 45 g. Oxygen left is 0.75 mol, or 24 g; hydrogen left is zero. Initial 69 g equals final 45 + 24 = 69 g.

A theoretical ceiling is not a measured yield

The calculated product assumes complete use of the limiting reagent in the specified reaction, pure starting material and no competing reactions. Actual recovered product can be smaller because reaction is incomplete, another product forms, material is lost during collection or the starting sample contains impurities. A result larger than the theoretical mass can indicate retained solvent, contamination, measurement error or incorrect assumptions. Neither discrepancy by itself identifies its cause.

At zero reactant supply, product is zero. When both supplies are zero, the displayed stoichiometric tie does not mean any reaction occurred. The app neither models ignition nor provides a physical mixing procedure. Save comparisons that change the limiting reagent, and use your reflection to propose measurements that would distinguish two explanations for a low recovered product mass.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Explain why floor rounding is inappropriate for moles. Diagnose a reported product mass that omits leftover reactant from the conservation ledger.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

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Previous: Connect a measured mass to a particle amount Next: Distinguish solution amount from concentration

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