Why Balanced Equation Coefficients Give Mole Ratios, Not Mass Ratios

Two beakers contain different chemicals, each made from exactly 5.00 g of dissolved solid. You mix them, and a white precipitate forms. The balanced equation has a 1:1 ratio between the reactants. Should that mean both chemicals are completely used up?

No. Balanced equation coefficients give mole ratios, not mass ratios, because chemical reactions depend on how many particles react, not how much those particles weigh. One mole of one substance can have a different mass from one mole of another.

That is why you cannot take a coefficient of 2 and automatically turn it into 2 g. You must convert the given mass into moles, use the coefficients to find the reacting amount, and then convert back to mass.

01What does a balanced equation actually count?

Consider the precipitation reaction between sodium carbonate and calcium chloride:

\[
\ce{Na2CO3(aq) + CaCl2(aq) -> CaCO3(s) + 2NaCl(aq)}
\]

The white precipitate is solid calcium carbonate, \(\ce{CaCO3}\).

Look at the coefficients:

• 1 unit of \(\ce{Na2CO3}\) reacts with 1 unit of \(\ce{CaCl2}\).
• This produces 1 unit of \(\ce{CaCO3}\) and the equivalent of 2 units of \(\ce{NaCl}\).

These are ratios of chemical amounts, not masses.

When sodium carbonate dissolves, it supplies sodium ions and carbonate ions. Calcium chloride supplies calcium ions and chloride ions.

The key change is:

\[
\ce{Ca^2+(aq) + CO3^2-(aq) -> CaCO3(s)}
\]

One calcium ion combines with one carbonate ion to form one formula unit of calcium carbonate.

The sodium and chloride ions remain dissolved in the water. The \(\ce{NaCl(aq)}\) in the full equation accounts for those ions; it does not mean separate sodium chloride molecules are floating around in solution.

Why does counting particles become counting moles?

A mole is a way of counting extremely large numbers of particles.

One mole contains approximately \(6.022 \times 10^{23}\) specified entities, such as atoms, molecules, ions, or formula units.

Imagine assembling bicycles. Each bicycle needs two wheels and one frame. That is a 2:1 component ratio, even though two wheels won’t have the same mass as one frame.

Chemical coefficients work similarly as counting ratios. The analogy stops at the assembly process: chemical reactions involve the rearrangement of atoms, ions, and bonds, not the mechanical attachment of ready-made parts.

For our reaction, the ratio is:

\[
\begin{aligned}
&n(\ce{Na2CO3}):n(\ce{CaCl2})\\
&\qquad =1:1
\end{aligned}
\]

Here, \(n\) means amount of substance, measured in moles (mol).

If 0.100 mol of sodium carbonate reacts completely, it consumes 0.100 mol of calcium chloride. If 0.500 mol reacts, it consumes 0.500 mol.

But that doesn’t mean the reactants have equal masses.

02An experiment that exposes the mistake

Suppose we want to test the prediction that equal masses of reactants should be completely consumed when their equation coefficients are equal.

We can design a precipitation experiment using the reaction above. The following quantities and outcomes are illustrative, not a report of measurements from an actual laboratory experiment.

The first design: mix the chemicals and weigh everything

Begin with two beakers.

The first contains 5.00 g of pure sodium carbonate dissolved in water. The second contains 5.00 g of pure, anhydrous calcium chloride dissolved in water.

The word anhydrous matters: it means the weighed calcium chloride contains no water of crystallisation.

Pour both solutions into a larger mixing beaker.

As the dissolved calcium ions encounter carbonate ions, calcium carbonate precipitates, turning the mixture cloudy white.

What should we expect?

A student who reads the 1:1 coefficients as a 1:1 mass ratio might predict that the two 5.00 g samples are exactly matched, so neither reactant remains.

But an important problem appears: simply watching the precipitate form does not tell us which reactant is left over.

Even measuring the total mass before and after mixing is not enough.

Assuming no material escapes, spills, or evaporates, the total mass remains unchanged. Both the correct mole-ratio model and the student’s incorrect mass-ratio prediction must obey conservation of mass.

We need a measurement that can distinguish them.

The revised design: isolate the product and test the remaining solution

After mixing, pour the suspension through a filter funnel lined with filter paper.

The solid calcium carbonate is retained on the paper. The liquid, called the filtrate, passes through into a receiving flask.

Dry the collected precipitate to constant mass and weigh it, accounting for the filter paper or container.

Then take a small separate sample of the filtrate and add a little more calcium chloride solution.

In an idealised trial using the stated masses, the calculated outcomes would be:

The additional precipitate is the crucial observation we would look for.

It indicates that carbonate ions remained after the original reaction. The two equal masses were not chemically equivalent amounts.

The dry precipitate mass also supports the mole-ratio calculation, provided the solid is sufficiently pure and recovery is complete.

In a real experiment, incomplete filtration, product loss, small amounts of dissolved calcium carbonate, and insufficient drying can all affect the measured mass. The experiment would need those sources of error controlled before its results could be used as strong quantitative evidence.

The balanced equation predicts the ideal relationship. The measurements test how closely the reaction follows that relationship.

03Equal moles can have very different masses

Let’s calculate what the coefficients actually imply.

The relevant molar masses are:

Molar mass is the mass of one mole of a substance.

Now imagine a reaction that consumes exactly 0.100 mol of each reactant.

Notice two things.

First, the coefficients of sodium carbonate and calcium chloride are both 1, but their masses differ: 10.60 g and 11.10 g.

Second, conservation of mass still works:

\[
\begin{aligned}
m_{\text{reactants}}
&=10.60+11.10\\
&=21.70\ \text{g}\\[4pt]
m_{\text{products}}
&=10.01+11.69\\
&=21.70\ \text{g}
\end{aligned}
\]

These masses refer to the reacting substances, excluding the unchanged solvent.

The equation doesn’t ignore mass. It conserves the atoms, which means total mass is conserved in a closed system.

It simply uses particle amounts to describe how the substances react.

Can a balanced equation give a mass ratio?

Yes, but not by using the coefficients alone.

For a complete reaction, the required masses of sodium carbonate and calcium chloride have the ratio:

\[
m(\ce{Na2CO3}):m(\ce{CaCl2})
=106.0:111.0
\]

That mass ratio comes from multiplying each mole amount by its molar mass.

It is not the coefficient ratio of 1:1.

The important distinction is that coefficients directly give mole ratios; molar masses are needed to calculate mass ratios.

04The reliable route: mass to moles to moles to mass

When a question gives the mass of one substance and asks for the mass of another, use the same sequence every time:

Given mass → moles of given substance → moles of required substance → required mass.

You need two relationships.

The first converts mass into amount:

\[
n=\frac{m}{M}
\]

Here:

• \(n\) is amount of substance in mol.
• \(m\) is mass in g.
• \(M\) is molar mass in g mol\(^{-1}\).

The second uses the balanced equation:

\[
n(B)=n(A)\frac{\nu_B}{\nu_A}
\]

Here, \(A\) is the starting substance, \(B\) is the substance being calculated, and \(\nu_A\) and \(\nu_B\) are their positive coefficients in the balanced equation.

For a reactant, this relationship concerns the amount actually consumed, not necessarily the amount initially present.

Finally, convert the calculated amount of \(B\) back into mass:

\[
m(B)=n(B)M(B)
\]

If you need more help with the first conversion, the Moles from Mass and Molar Mass HSC Chemistry guide covers it in detail.

05How to recognise the mass-ratio mistake

There are three especially useful warning signs.

Mistake 1: Using coefficients directly on grams

Suppose an equation has a 2:1 ratio.

You cannot conclude that 2 g of one substance reacts with 1 g of the other.

The 2:1 relationship applies to moles. To obtain a mass ratio, multiply each coefficient by the corresponding molar mass.

Mistake 2: Multiplying by the molar mass before using the mole ratio

If your starting measurement is in grams, first divide by molar mass.

Use the coefficient ratio only after reaching moles.

Then multiply by the required substance’s molar mass to get grams again.

A useful unit check is:

\[
\text{g}
\longrightarrow\text{mol}
\longrightarrow\text{mol}
\longrightarrow\text{g}
\]

Mistake 3: Assuming a balanced equation means all reactants disappear

A balanced equation gives the proportions in which substances react. It does not promise that the starting mixture contains those proportions.

If two reactants are supplied in the wrong mole ratio, one may run out while the other remains.

That remaining substance is not evidence that the coefficients are incorrect.

It is evidence that the reactants were supplied in non-stoichiometric amounts.

06Questions and solutions

Question 1

Magnesium reacts with oxygen according to:

\[
\ce{2Mg(s) + O2(g) -> 2MgO(s)}
\]

A 4.86 g sample of magnesium reacts completely with excess oxygen.

Calculate the theoretical mass of magnesium oxide formed and the mass of oxygen consumed.

Use \(M(\ce{Mg})=24.3\ \text{g mol}^{-1}\) and \(M(\ce{O})=16.0\ \text{g mol}^{-1}\).

Solution 1

The reaction produces 8.06 g of magnesium oxide and consumes 3.20 g of oxygen.

The coefficients show a 2:2 mole ratio between magnesium and magnesium oxide, so their reacting mole amounts are equal.

First, calculate magnesium’s amount:

\[
\begin{aligned}
n(\ce{Mg})
&=\frac{4.86}{24.3}\\
&=0.200\ \text{mol}
\end{aligned}
\]

Therefore, \(n(\ce{MgO})=0.200\ \text{mol}\).

The molar mass of magnesium oxide is \(24.3+16.0=40.3\ \text{g mol}^{-1}\).

\[
\begin{aligned}
m(\ce{MgO})
&=0.200\times40.3\\
&=8.06\ \text{g}
\end{aligned}
\]

From the 2:1 ratio of magnesium to oxygen:

\[
\begin{aligned}
n(\ce{O2})
&=0.200\times\frac12\\
&=0.100\ \text{mol}\\
m(\ce{O2})
&=0.100\times32.0\\
&=3.20\ \text{g}
\end{aligned}
\]

The mass check is \(4.86+3.20=8.06\ \text{g}\).

The tempting mistake is to assume that because magnesium and magnesium oxide have equal coefficients, they must have equal masses. Their mole amounts are equal, but the oxide includes oxygen atoms that were not part of the original magnesium sample.

Question 2

A student mixes solutions containing 2.12 g of pure sodium carbonate and 3.33 g of pure, anhydrous calcium chloride.

The balanced reaction is:

\[
\ce{Na2CO3 + CaCl2 -> CaCO3 + 2NaCl}
\]

Assume complete precipitation and no material loss.

(a) Calculate the theoretical dry mass of calcium carbonate.

(b) Calculate the mass of the excess reactant remaining.

(c) Explain how the filtrate could be tested to identify the remaining reactant.

(d) Another student argues that the isolated precipitate should weigh 5.45 g because total mass is conserved. Explain the error.

Use the molar masses from the earlier table.

Solution 2

Sodium carbonate is limiting, 2.00 g of calcium carbonate forms, and 1.11 g of calcium chloride remains. The filtrate can be tested for the remaining calcium ions.

(a) Calculate the initial amounts:

\[
\begin{aligned}
n(\ce{Na2CO3})
&=\frac{2.12}{106.0}\\
&=0.0200\ \text{mol}\\[4pt]
n(\ce{CaCl2})
&=\frac{3.33}{111.0}\\
&=0.0300\ \text{mol}
\end{aligned}
\]

The required ratio is 1:1. Sodium carbonate runs out first, so only 0.0200 mol of calcium carbonate forms.

\[
\begin{aligned}
m(\ce{CaCO3})
&=0.0200\times100.1\\
&=2.00\ \text{g}
\end{aligned}
\]

(b) The calcium chloride remaining is:

\[
\begin{aligned}
n_{\text{left}}
&=0.0300-0.0200\\
&=0.0100\ \text{mol}\\
m_{\text{left}}
&=0.0100\times111.0\\
&=1.11\ \text{g}
\end{aligned}
\]

(c) Add a small amount of sodium carbonate solution to a separate sample of the filtrate.

A new white calcium carbonate precipitate should form because calcium ions remain in solution.

(d) The 5.45 g total includes the substances that remain dissolved, not just the collected solid.

The reaction also produces 0.0400 mol of sodium chloride, corresponding to approximately 2.34 g.

Accounting for the reaction products and leftover reactant gives:

\[
2.00+2.34+1.11=5.45\ \text{g}
\]

The mistaken argument applies conservation of mass to one isolated product rather than the entire system.

Question 3

A student heats 6.35 g of copper in an oxygen atmosphere. After the reaction, the solid product has a mass of 7.15 g.

Two possible models are proposed:

Model A:

\[
\ce{4Cu + O2 -> 2Cu2O}
\]

Model B:

\[
\ce{2Cu + O2 -> 2CuO}
\]

Assume that all the copper reacts, only one of these oxides forms, and no solid is lost.

(a) Determine which model is consistent with the observed masses.

(b) Calculate the amount of oxygen gas consumed.

(c) Explain why the increase in the solid’s mass does not violate conservation of mass.

Use \(M(\ce{Cu})=63.5\ \text{g mol}^{-1}\) and \(M(\ce{O})=16.0\ \text{g mol}^{-1}\).

Solution 3

Model A, forming \(\ce{Cu2O}\), is consistent with the measurements. The copper has incorporated 0.800 g of oxygen, equivalent to 0.0250 mol of \(\ce{O2}\).

(a) The oxygen mass incorporated is:

\[
\begin{aligned}
m(\ce{O2})
&=7.15-6.35\\
&=0.800\ \text{g}
\end{aligned}
\]

Calculate the amount of copper initially present:

\[
\begin{aligned}
n(\ce{Cu})
&=\frac{6.35}{63.5}\\
&=0.100\ \text{mol}
\end{aligned}
\]

(b) The oxygen amount is:

\[
\begin{aligned}
n(\ce{O2})
&=\frac{0.800}{32.0}\\
&=0.0250\ \text{mol}
\end{aligned}
\]

The measured reacting mole ratio is:

\[
n(\ce{Cu}):n(\ce{O2})
=0.100:0.0250=4:1
\]

Model A requires exactly that 4:1 ratio.

Model B instead requires a 2:1 ratio and would consume 0.0500 mol of oxygen for 0.100 mol of copper. That would add 1.60 g of oxygen and produce 7.95 g of oxide, not 7.15 g.

(c) The solid gains mass because oxygen from the surrounding atmosphere becomes part of it.

The original 6.35 g figure includes copper only. The final figure includes copper and oxygen.

The tempting interpretation is that the solid has somehow gained unexplained mass. The missing part of that argument is the gaseous reactant entering the solid. With oxygen accounted for, mass is conserved.

The conclusion depends on the stated assumption that the product is a pure sample of one of the two proposed oxides. A mixture of products or incomplete conversion would require more evidence.

Question 4

An unlabelled white solid is known to be either:

• Anhydrous calcium chloride, \(\ce{CaCl2}\), with molar mass 111.0 g mol\(^{-1}\).
• Calcium chloride dihydrate, \(\mathrm{CaCl_2\cdot 2H_2O}\), with molar mass 147.0 g mol\(^{-1}\).

Both dissolve to supply calcium ions.

A 2.94 g sample of the unknown solid is dissolved in water. The student adds 2.12 g of pure sodium carbonate, also dissolved in water, and collects 2.00 g of dry calcium carbonate.

Assume complete precipitation, no side reactions, and perfect recovery.

(a) Is the 2.00 g result enough to determine which calcium chloride compound was present?

(b) The student repeats the experiment using a fresh 2.94 g sample of the unknown, but increases the sodium carbonate to 3.18 g. Calculate the theoretical precipitate mass for each possible identity.

(c) Explain why the different masses do not mean the 1:1 mole ratio has changed.

Use \(M(\ce{Na2CO3})=106.0\ \text{g mol}^{-1}\) and \(M(\ce{CaCO3})=100.1\ \text{g mol}^{-1}\).

Solution 4

No, the first experiment cannot distinguish the two substances. The revised experiment can: it predicts 2.65 g of precipitate for anhydrous calcium chloride and 2.00 g for the dihydrate.

(a) The first experiment uses:

\[
\begin{aligned}
n(\ce{Na2CO3})
&=\frac{2.12}{106.0}\\
&=0.0200\ \text{mol}
\end{aligned}
\]

If the unknown is anhydrous calcium chloride:

\[
\begin{aligned}
n(\ce{CaCl2})
&=\frac{2.94}{111.0}\\
&=0.0265\ \text{mol}
\end{aligned}
\]

There is excess calcium chloride, so sodium carbonate limits the reaction to 0.0200 mol of precipitate.

If the unknown is the dihydrate:

\[
\begin{aligned}
n(\text{dihydrate})
&=\frac{2.94}{147.0}\\
&=0.0200\ \text{mol}
\end{aligned}
\]

That also supplies 0.0200 mol of calcium ions.

Both cases produce:

\[
\begin{aligned}
m(\ce{CaCO3})
&=0.0200\times100.1\\
&=2.00\ \text{g}
\end{aligned}
\]

The student might be tempted to identify the dihydrate because its amount matches the amount of sodium carbonate exactly.

But the anhydrous sample also gives the same precipitate mass because sodium carbonate is limiting.

(b) In the revised experiment:

\[
\begin{aligned}
n(\ce{Na2CO3})
&=\frac{3.18}{106.0}\\
&=0.0300\ \text{mol}
\end{aligned}
\]

Now sodium carbonate is in excess for both possible calcium chloride compounds.

For anhydrous calcium chloride:

\[
\begin{aligned}
m(\ce{CaCO3})
&=\frac{2.94}{111.0}\times100.1\\
&=2.65\ \text{g}
\end{aligned}
\]

For calcium chloride dihydrate:

\[
\begin{aligned}
m(\ce{CaCO3})
&=\frac{2.94}{147.0}\times100.1\\
&=2.00\ \text{g}
\end{aligned}
\]

The revised design makes the amount of calcium-containing substance determine the product mass, rather than allowing sodium carbonate to limit both possibilities to the same result.

(c) The calcium-ion-to-carbonate-ion mole ratio is still 1:1.

The hydrate contains water of crystallisation, increasing the mass associated with one mole of calcium chloride units.

Consequently, equal sample masses supply different mole amounts of calcium ions.

The deeper experimental lesson is that two competing models can produce identical results under one limiting condition. A useful revision changes that condition so their predictions separate.

Question 5

A 0.2565 g mixture contains only magnesium and aluminium metals.

The metals react with hydrochloric acid according to:

\[
\ce{Mg + 2HCl -> MgCl2 + H2}
\]

\[
\ce{2Al + 6HCl -> 2AlCl3 + 3H2}
\]

A gas-tight flask is connected by a delivery tube to a gas syringe. Both metals are placed in the flask.

In Stage 1, exactly 0.0160 mol of hydrochloric acid is introduced. When gas production stops, the syringe contains 0.192 L of hydrogen.

In Stage 2, more hydrochloric acid is added in excess. All the remaining metal reacts, and the total hydrogen collected across both stages is 0.300 L.

Assume that both metals can react, only the stated reactions occur, the reaction gases are fully collected, and the same temperature and pressure are maintained. Under these measurement conditions, 1.00 mol of hydrogen occupies 24.0 L.

No assumption is made about which metal reacts faster or first.

(a) Determine the original amounts, in moles, of magnesium and aluminium.

(b) Show that all the hydrochloric acid introduced in Stage 1 was consumed.

(c) Can the mass of unreacted metal remaining at the end of Stage 1 be determined uniquely? If not, determine its possible minimum and maximum values, assuming the stated reactions and allowing either metal to react to any extent.

(d) A student argues that the first hydrogen volume must reveal the original metal mixture because aluminium produces more hydrogen per mole than magnesium. Explain the flaw in that argument.

Use \(M(\ce{Mg})=24.3\ \text{g mol}^{-1}\) and \(M(\ce{Al})=27.0\ \text{g mol}^{-1}\).

Solution 5

The original mixture contained 0.00500 mol of magnesium and 0.00500 mol of aluminium. All Stage 1 acid was consumed, but the mass of metal remaining after that stage is not uniquely determined: it can range from 0.0810 g to approximately 0.109 g.

The key is that both metals consume two moles of hydrochloric acid for every mole of hydrogen produced, despite producing different amounts of hydrogen per mole of metal.

(a) First use the total hydrogen volume after all metal has reacted:

\[
\begin{aligned}
n(\ce{H2})
&=\frac{0.300}{24.0}\\
&=0.01250\ \text{mol}
\end{aligned}
\]

Let \(x\) be the initial amount of magnesium in mol and \(y\) the initial amount of aluminium in mol.

From the equations, 1 mol of magnesium produces 1 mol of hydrogen, while 1 mol of aluminium produces 1.5 mol of hydrogen.

Therefore:

\[
x+1.5y=0.01250
\]

The total metal mass provides another equation:

\[
24.3x+27.0y=0.2565
\]

The molar masses in this expression are in g mol\(^{-1}\), so each term represents a mass in grams.

Solving simultaneously:

\[
\begin{aligned}
x&=0.01250-1.5y\\
0.2565&=0.30375-9.45y\\
y&=0.00500\ \text{mol}\\
x&=0.00500\ \text{mol}
\end{aligned}
\]

Thus the initial masses were:

\[
\begin{aligned}
m(\ce{Mg})
&=0.00500\times24.3\\
&=0.1215\ \text{g}\\[4pt]
m(\ce{Al})
&=0.00500\times27.0\\
&=0.1350\ \text{g}
\end{aligned}
\]

These add to 0.2565 g, as required.

(b) In Stage 1:

\[
\begin{aligned}
n(\ce{H2})
&=\frac{0.192}{24.0}\\
&=0.00800\ \text{mol}
\end{aligned}
\]

Both reaction equations show that two moles of HCl are consumed for each mole of hydrogen formed.

Therefore:

\[
\begin{aligned}
n(\ce{HCl})_{\text{used}}
&=2(0.00800)\\
&=0.0160\ \text{mol}
\end{aligned}
\]

This equals the entire initial acid amount, so Stage 1 consumed all the supplied hydrochloric acid.

There was also metal left because the completed reaction eventually produced 0.01250 mol of hydrogen, more than the 0.00800 mol formed during Stage 1.

(c) Here is the difficult part.

Knowing the original composition does not tell us which metal was consumed during Stage 1.

Consider two possible reaction histories.

Possibility A: All magnesium reacts before all aluminium has reacted.

The 0.00500 mol magnesium produces 0.00500 mol hydrogen.

The remaining Stage 1 hydrogen comes from aluminium:

\[
\begin{aligned}
n(\ce{H2})_{\text{from Al}}
&=0.00800-0.00500\\
&=0.00300\ \text{mol}
\end{aligned}
\]

Since 1 mol of aluminium produces 1.5 mol hydrogen, the aluminium consumed is:

\[
\begin{aligned}
n(\ce{Al})_{\text{used}}
&=\frac{0.00300}{1.5}\\
&=0.00200\ \text{mol}
\end{aligned}
\]

The remaining aluminium is \(0.00500-0.00200=0.00300\ \text{mol}\).

Its mass is:

\[
m_{\text{remaining}}
=0.00300\times27.0
=0.0810\ \text{g}
\]

Possibility B: All aluminium reacts before all magnesium has reacted.

The 0.00500 mol aluminium produces:

\[
\begin{aligned}
n(\ce{H2})_{\text{from Al}}
&=1.5(0.00500)\\
&=0.00750\ \text{mol}
\end{aligned}
\]

The remaining hydrogen amount, \(0.00800-0.00750=0.00050\ \text{mol}\), comes from magnesium.

Therefore, 0.00050 mol magnesium reacts, leaving:

\[
\begin{aligned}
n(\ce{Mg})_{\text{left}}
&=0.00500-0.00050\\
&=0.00450\ \text{mol}
\end{aligned}
\]

The remaining metal mass is:

\[
\begin{aligned}
m_{\text{remaining}}
&=0.00450\times24.3\\
&=0.10935\ \text{g}
\end{aligned}
\]

Both histories satisfy the experimental data and the balanced equations.

They give the minimum and maximum possible residue masses. Intermediate distributions of reaction between the two metals give intermediate residue masses.

So the Stage 1 mass cannot be uniquely determined without additional information about the individual reaction extents.

(d) The student’s argument overlooks which reagent is limiting the first stage.

Aluminium does produce more hydrogen per mole of metal than magnesium:

\[
1\ \text{mol Al}\longrightarrow1.5\ \text{mol H2}
\]

compared with:

\[
1\ \text{mol Mg}\longrightarrow1\ \text{mol H2}
\]

But both reactions require the same amount of hydrochloric acid for each mole of hydrogen produced:

\[
2\ \text{mol HCl}\longrightarrow1\ \text{mol H2}
\]

Because Stage 1 is acid-limited, its final hydrogen amount is fixed by the acid supply, not by the particular combination of metals that reacted.

Only after sufficient additional acid has been supplied to consume both metals does the total hydrogen amount, together with the initial total metal mass, determine their original proportions.

That is the conceptual trap: different reactions can have different metal-to-hydrogen mole ratios while sharing the same acid-to-hydrogen mole ratio. Which ratio controls the measurement depends on the limiting condition.

07What this lets you work out next

A balanced equation is a statement about the amounts of substances consumed and produced. Mass enters when those amounts are converted using molar masses.

That gives you a reliable way to approach reaction-mass calculations:

Convert the supplied mass to moles, identify the limiting reagent if necessary, apply the coefficient ratio, and convert the required amount back to mass.

The next useful question is how you can establish a mole ratio when the balanced equation is not given. That requires experimental evidence, often collected by changing the amount of one reactant and measuring when the product stops increasing.

The Whole-number mole ratios from experimental data guide develops that measurement problem, including how to recognise the point where a limiting reagent changes.

That connects the two directions of quantitative chemistry: using a known equation to predict measurements, and using measurements to test the proportions represented by an equation.