How to Identify a Limiting Reagent from a Product-Mass Graph

A product-mass graph rises as more of one reactant is added, then becomes horizontal. Why does adding more reactant suddenly stop producing more product?

The turning point tells you when the limiting reagent changes. Before the turning point, the reactant being added is limiting. After it, the reactant held constant is limiting. At the turning point, both reactants are present in exactly the required mole ratio.

To identify the limiting reagent from a graph, use this three-step approach:

1. Locate the breakpoint, where the product mass stops increasing.
2. Convert the reactant quantities into moles and divide each amount by its balanced equation coefficient.
3. Calculate the theoretical product mass and the amount of excess reactant remaining.

The graph gives you the pattern. The balanced equation proves why that pattern occurs.

01Why does a product-mass graph rise and then level off?

Consider an investigation where magnesium chloride solution reacts with sodium hydroxide solution to produce a white precipitate of magnesium hydroxide.

You keep the amount of magnesium chloride constant and gradually increase the amount of sodium hydroxide.

Initially, more sodium hydroxide produces more magnesium hydroxide. Eventually, however, adding extra sodium hydroxide makes no difference to the mass of precipitate.

Before calculating anything, predict which reactant limits the reaction after the graph becomes horizontal.

It is magnesium chloride. Once all the available magnesium ions have reacted, there is nothing left for the additional hydroxide ions to react with.

Understanding the reaction

The balanced chemical equation is:

\[ \ce{MgCl2(aq) + 2NaOH(aq) -> Mg(OH)2(s) + 2NaCl(aq)} \]

The coefficients tell us that:

• 1 mol of magnesium chloride reacts with 2 mol of sodium hydroxide.
• 1 mol of magnesium hydroxide forms for every 1 mol of magnesium chloride consumed.
• Sodium chloride remains dissolved in the solution.

The solid magnesium hydroxide is the product we collect and weigh.

How would we investigate this experimentally?

Imagine starting with:

• 50.0 mL of 1.00 mol L⁻¹ magnesium chloride solution.
• 1.00 mol L⁻¹ sodium hydroxide solution, added in different volumes.

For each trial, the solutions are mixed, and the magnesium hydroxide precipitate is allowed to form.

The solid is then filtered, washed to remove soluble substances, and dried to constant mass. Its mass is determined by subtracting the mass of the filter paper from the combined mass of the filter paper and precipitate.

A single trial would tell us how much precipitate forms at one particular volume of sodium hydroxide. It would not tell us where the limiting reagent changes.

A better investigation uses several different sodium hydroxide volumes, including values below and above the predicted breakpoint.

Keep the magnesium chloride quantity, both solution concentrations, and experimental conditions consistent. Repeat measurements to improve reliability.

The graph we develop below uses illustrative theoretical data, rather than claiming to report measurements from an actual experiment.

What should happen as more sodium hydroxide is added?

At the beginning, there is plenty of magnesium chloride but relatively little sodium hydroxide.

As sodium hydroxide is added, more hydroxide ions react with magnesium ions. More magnesium hydroxide precipitates, so the product mass increases.

Eventually, all the available magnesium chloride has reacted.

Adding more sodium hydroxide after this point cannot produce additional magnesium hydroxide. The product mass stays constant.

This gives us the characteristic shape:

• An increasing section, where sodium hydroxide is limiting.
• A breakpoint, where both reactants are present in the exact stoichiometric ratio.
• A horizontal section, where magnesium chloride is limiting and sodium hydroxide is in excess.

02How to identify the limiting reagent from the graph

Let’s work through the data rather than relying on the shape alone.

We have a fixed amount of magnesium chloride:

\[ n=cV \]

Here:

• \(n\) is the amount of substance in mol.
• \(c\) is the molar concentration in mol L\(^{-1}\).
• \(V\) is the solution volume in L.

Therefore:

\[ n(\ce{MgCl2})=1.00\times0.0500 \]

\[ =0.0500\text{ mol} \]

The magnesium chloride amount stays at 0.0500 mol throughout the investigation.

The sodium hydroxide concentration is also fixed at 1.00 mol L\(^{-1}\), but its volume changes.

Illustrative data

The following table shows the predicted theoretical mass of magnesium hydroxide for different sodium hydroxide volumes.

The theoretical product mass increases until 100.0 mL of sodium hydroxide has been added.

After that, it stays at approximately 2.92 g.

Step 1: Find the breakpoint

The breakpoint occurs at:

\[ V(\ce{NaOH})=100.0\text{ mL} \]

Convert this to litres:

\[ V=0.1000\text{ L} \]

Calculate the amount of sodium hydroxide:

\[ n=cV \]

\[ =1.00\times0.1000 \]

\[ =0.1000\text{ mol} \]

At this point, we have:

• 0.0500 mol of magnesium chloride.
• 0.1000 mol of sodium hydroxide.

These amounts are in the exact 1:2 ratio required by the balanced equation.

Neither reactant is in excess at the breakpoint.

Step 2: Understand what happens on either side

The most important feature is the change in limiting reagent.

Why?

Before the breakpoint, there is insufficient sodium hydroxide to consume all the magnesium chloride.

After the breakpoint, there is more sodium hydroxide than the magnesium chloride can react with. Magnesium chloride is completely consumed first.

At the breakpoint, both reactants are completely consumed, assuming the reaction proceeds to completion.

03The mathematical method: divide by the coefficient

The graph helps you identify the likely limiting reagent, but there is a reliable calculation to confirm your answer.

For a general balanced equation:

\[ \ce{aA + bB -> products} \]

Calculate:

\[ \frac{n(A)}{a} \qquad\text{and}\qquad \frac{n(B)}{b} \]

The smaller result identifies the limiting reagent.

Why does this work?

Each calculation tells you how many complete reaction units that reactant can support.

The reactant that supports fewer reaction units runs out first.

For the magnesium chloride reaction:

\[ \ce{MgCl2 + 2NaOH -> Mg(OH)2 + 2NaCl} \]

Compare:

\[ \frac{n(\ce{MgCl2})}{1} \qquad\text{and}\qquad \frac{n(\ce{NaOH})}{2} \]

The smaller value identifies the limiting reagent.

This is especially useful when the reactants have different coefficients. Simply comparing their mole amounts can give you the wrong answer.

If you need a refresher on this calculation, see How to Compare Reactant Amounts and Find the Limiting Reactant.

What happens to the excess reactant?

Identifying the limiting reagent is only part of the calculation.

You also need to calculate how much of the excess reactant remains.

Once the limiting reagent is identified:

1. Calculate how much of it reacts.
2. Use the balanced equation to calculate how much excess reactant is consumed.
3. Subtract the amount consumed from the initial amount.

The general relationship is:

\[ n_{\text{left}}= n_{\text{initial}}-n_{\text{consumed}} \]

You can then calculate the theoretical product mass using the limiting reagent.

Remember:

\[ m=nM \]

where:

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

This is the same stoichiometric method used in ordinary limiting-reagent calculations. The graph simply gives you another way to recognise where the limiting reagent changes.

06Three mistakes to avoid when interpreting product-mass graphs

1. Assuming the reactant on the horizontal axis is always limiting

The reactant being added is limiting only before the breakpoint.

After the breakpoint, the fixed reactant is limiting.

For our example:

• At 50.0 mL NaOH, sodium hydroxide is limiting.
• At 100.0 mL NaOH, neither reactant is in excess.
• At 150.0 mL NaOH, magnesium chloride is limiting.

2. Comparing masses or mole amounts without considering coefficients

A reaction might require two moles of one reactant for every mole of another.

The reactant with fewer moles is not automatically limiting.

Always compare:

\[ \frac{\text{amount of reactant}} {\text{balanced equation coefficient}} \]

3. Assuming a lower product mass means a different limiting reagent

A measured product mass can be lower than the theoretical prediction because of:

• Product lost during filtration or transfer.
• Incomplete reaction.
• Product remaining dissolved in solution.
• Other experimental limitations.

A lower measured mass does not automatically mean the limiting reagent has changed.

The graph’s horizontal position and breakpoint help identify the stoichiometric relationship. The vertical product mass helps determine the yield.

One important limitation: the usual rise-and-plateau model assumes that the excess reactant does not consume the product. Some reactions behave differently, as we’ll explore in the final question.

07Questions and solutions

Question 1

A student investigates the reaction between magnesium chloride and sodium hydroxide.

The balanced equation is:

\[ \ce{MgCl2 + 2NaOH -> Mg(OH)2 + 2NaCl} \]

The student keeps 40.0 mL of 1.00 mol L⁻¹ magnesium chloride solution constant and adds 1.00 mol L⁻¹ sodium hydroxide.

An idealised product-mass graph has a breakpoint at 80.0 mL of sodium hydroxide.

At 60.0 mL of sodium hydroxide, the theoretical product mass is approximately 1.75 g.

Calculate:

a) The limiting reagent at 60.0 mL.

b) The theoretical mass of magnesium hydroxide produced.

c) The amount of magnesium chloride remaining.

Use \(M(\ce{Mg(OH)2})=58.3\text{ g mol}^{-1}\).

Solution 1

Sodium hydroxide is limiting at 60.0 mL. The theoretical magnesium hydroxide mass is 1.75 g, and 0.0100 mol of magnesium chloride remains.

Step 1: Calculate the initial moles of each reactant.

Magnesium chloride:

\[ n=1.00\times0.0400 \]

\[ =0.0400\text{ mol} \]

Sodium hydroxide:

\[ n=1.00\times0.0600 \]

\[ =0.0600\text{ mol} \]

Step 2: Divide by the coefficients.

\[ \ce{MgCl2}:\quad\frac{0.0400}{1}=0.0400 \]

\[ \ce{NaOH}:\quad\frac{0.0600}{2}=0.0300 \]

Sodium hydroxide has the smaller value, so it is limiting.

Step 3: Calculate the amount of product.

The equation gives a 2:1 ratio between NaOH and Mg(OH)₂.

\[ n(\ce{Mg(OH)2})=\frac{0.0600}{2} \]

\[ =0.0300\text{ mol} \]

Step 4: Calculate the theoretical product mass.

\[ m=nM \]

\[ =0.0300\times58.3 \]

\[ \boxed{1.75\text{ g}} \]

Step 5: Calculate the magnesium chloride remaining.

One mole of magnesium chloride reacts for every two moles of sodium hydroxide.

Therefore:

\[ n(\ce{MgCl2})_{\text{consumed}} =0.0300\text{ mol} \]

\[ n(\ce{MgCl2})_{\text{left}} =0.0400-0.0300 \]

\[ \boxed{0.0100\text{ mol}} \]

The answer agrees with the graph: 60.0 mL is before the 80.0 mL breakpoint, so sodium hydroxide is limiting.

Question 2

A second investigation uses 40.0 mL of 1.00 mol L⁻¹ magnesium chloride solution and 120.0 mL of 1.00 mol L⁻¹ sodium hydroxide solution.

After filtration and drying, the student collects 2.20 g of magnesium hydroxide.

Assume the reaction proceeds to completion and the difference between the theoretical and collected product mass is caused only by product loss during filtration.

Calculate:

a) The limiting reagent.

b) The amount of excess reactant remaining.

c) The theoretical mass of magnesium hydroxide.

d) The percentage yield.

Explain why the actual product mass does not change the identity of the limiting reagent.

Solution 2

Magnesium chloride is limiting. The reaction leaves 0.0400 mol of sodium hydroxide, produces a theoretical 2.33 g of magnesium hydroxide, and has a percentage yield of 94.3%.

Step 1: Calculate the initial mole amounts.

For magnesium chloride:

\[ n=1.00\times0.0400 \]

\[ =0.0400\text{ mol} \]

For sodium hydroxide:

\[ n=1.00\times0.1200 \]

\[ =0.1200\text{ mol} \]

Step 2: Identify the limiting reagent.

Divide each amount by its coefficient:

\[ \ce{MgCl2}:\quad\frac{0.0400}{1}=0.0400 \]

\[ \ce{NaOH}:\quad\frac{0.1200}{2}=0.0600 \]

Magnesium chloride has the smaller value, so it is limiting.

Step 3: Calculate the sodium hydroxide consumed.

\[ n(\ce{NaOH})_{\text{consumed}} =2\times0.0400 \]

\[ =0.0800\text{ mol} \]

Therefore:

\[ n(\ce{NaOH})_{\text{left}} =0.1200-0.0800 \]

\[ \boxed{0.0400\text{ mol}} \]

Step 4: Calculate the theoretical product mass.

One mole of magnesium chloride produces one mole of magnesium hydroxide.

\[ n(\ce{Mg(OH)2})=0.0400\text{ mol} \]

\[ m=nM \]

\[ =0.0400\times58.3 \]

\[ \boxed{2.332\text{ g}} \]

Step 5: Calculate the percentage yield.

\[ \%\text{ yield} =\frac{2.20}{2.332}\times100 \]

\[ \boxed{94.3\%} \]

The actual product mass is below the theoretical maximum because some precipitate is lost during filtration.

This does not mean sodium hydroxide has become limiting. The balanced equation and initial amounts still establish magnesium chloride as the limiting reagent.

Question 3

A student investigates the reaction:

\[ \ce{MgCl2 + 2NaOH -> Mg(OH)2 + 2NaCl} \]

The student uses 50.0 mL of magnesium chloride solution with an unknown concentration and adds 0.600 mol L⁻¹ sodium hydroxide.

The resulting product-mass graph has:

• A breakpoint at 80.0 mL of sodium hydroxide.
• A measured horizontal plateau at 1.26 g of magnesium hydroxide.

A separate recovery test establishes that 90.0% of the magnesium hydroxide is collected consistently at every sodium hydroxide volume.

Calculate:

a) The initial amount of magnesium chloride.

b) The initial concentration of magnesium chloride.

c) The theoretical maximum mass of magnesium hydroxide.

d) Explain why using the measured plateau directly to calculate the initial magnesium chloride amount gives an incorrect answer.

Solution 3

The initial magnesium chloride amount is 0.0240 mol, its concentration is 0.480 mol L⁻¹, and the theoretical maximum magnesium hydroxide mass is 1.40 g.

The measured plateau is lower because only 90.0% of the precipitate is recovered.

Step 1: Use the breakpoint to calculate the amount of sodium hydroxide.

Convert the volume into litres:

\[ V=80.0\div1000 \]

\[ =0.0800\text{ L} \]

Calculate the moles:

\[ n=cV \]

\[ =0.600\times0.0800 \]

\[ =0.0480\text{ mol} \]

Step 2: Use the balanced equation to determine the magnesium chloride amount.

At the breakpoint, both reactants are present in their exact stoichiometric ratio.

The equation requires two moles of NaOH for every mole of MgCl₂.

Therefore:

\[ n(\ce{MgCl2})=\frac{0.0480}{2} \]

\[ \boxed{0.0240\text{ mol}} \]

Step 3: Calculate the magnesium chloride concentration.

The original volume is 50.0 mL, or 0.0500 L.

\[ c=\frac{n}{V} \]

\[ =\frac{0.0240}{0.0500} \]

\[ \boxed{0.480\text{ mol L}^{-1}} \]

Step 4: Calculate the theoretical maximum product mass.

The equation produces one mole of Mg(OH)₂ for every mole of MgCl₂.

\[ n(\ce{Mg(OH)2})=0.0240\text{ mol} \]

\[ m=nM \]

\[ =0.0240\times58.3 \]

\[ =1.3992\text{ g} \]

\[ \boxed{1.40\text{ g}} \]

Step 5: Check the measured plateau.

The collection efficiency is 90.0%.

Therefore:

\[ m_{\text{collected}} =1.3992\times0.900 \]

\[ =1.2593\text{ g} \]

\[ \boxed{1.26\text{ g}} \]

This agrees with the graph.

The tempting approach is to divide 1.26 g directly by the molar mass of magnesium hydroxide:

\[ n=\frac{1.26}{58.3} \]

\[ =0.0216\text{ mol} \]

However, this is the amount recovered, not the theoretical amount produced.

The breakpoint gives the original stoichiometric amount. The recovery efficiency explains the lower measured plateau.

This distinction is essential when interpreting experimental graphs. A low product mass does not automatically mean that less limiting reagent was initially present.

Question 4

Two students investigate the same reaction:

\[ \ce{MgCl2 + 2NaOH -> Mg(OH)2 + 2NaCl} \]

Both start with 0.0500 mol of magnesium chloride.

Student A uses 1.00 mol L⁻¹ sodium hydroxide and observes a breakpoint at 100 mL.

Student B uses 0.500 mol L⁻¹ sodium hydroxide and observes a breakpoint at 200 mL.

Both graphs reach the same theoretical plateau of 2.92 g.

A student argues that Student B must have started with twice as much magnesium chloride because the breakpoint occurs at twice the volume.

Evaluate this claim.

Then, for Student B, calculate at an added sodium hydroxide volume of 150 mL:

a) The limiting reagent.

b) The theoretical mass of magnesium hydroxide.

c) The amount of magnesium chloride remaining.

Solution 4

The claim is incorrect. Both students started with the same amount of magnesium chloride, and both breakpoints correspond to 0.100 mol of sodium hydroxide.

Student B simply used a solution with half the concentration.

Step 1: Compare the amounts of sodium hydroxide at each breakpoint.

For Student A:

\[ n=cV \]

\[ =1.00\times0.100 \]

\[ =0.100\text{ mol} \]

For Student B:

\[ n=cV \]

\[ =0.500\times0.200 \]

\[ =0.100\text{ mol} \]

Both students have added the same amount of sodium hydroxide at their respective breakpoints.

Since each has 0.0500 mol of magnesium chloride, both mixtures are stoichiometric:

\[ \frac{0.0500}{1} = \frac{0.100}{2} \]

The difference in breakpoint volume is caused by the difference in concentration, not the initial amount of magnesium chloride.

Step 2: Calculate the amount of sodium hydroxide added by Student B at 150 mL.

Convert the volume into litres:

\[ V=0.150\text{ L} \]

\[ n(\ce{NaOH})=0.500\times0.150 \]

\[ =0.0750\text{ mol} \]

Step 3: Identify the limiting reagent.

Divide each reactant amount by its coefficient:

\[ \ce{MgCl2}:\quad\frac{0.0500}{1}=0.0500 \]

\[ \ce{NaOH}:\quad\frac{0.0750}{2}=0.0375 \]

Sodium hydroxide is limiting.

Step 4: Calculate the theoretical product mass.

\[ n(\ce{Mg(OH)2})=\frac{0.0750}{2} \]

\[ =0.0375\text{ mol} \]

Convert into mass:

\[ m=nM \]

\[ =0.0375\times58.3 \]

\[ \boxed{2.19\text{ g}} \]

Step 5: Calculate the magnesium chloride remaining.

The amount consumed is:

\[ n(\ce{MgCl2})_{\text{consumed}} =0.0375\text{ mol} \]

Therefore:

\[ n(\ce{MgCl2})_{\text{left}} =0.0500-0.0375 \]

\[ \boxed{0.0125\text{ mol}} \]

The key distinction is that volume alone does not tell you the amount of reactant present.

When the concentration changes, the breakpoint can move horizontally even though the initial quantity of the fixed reactant stays exactly the same.

Question 5

Extension challenge: When does the product-mass graph stop following the usual plateau pattern?

A student investigates the reaction between aluminium chloride and sodium hydroxide.

The initial amount of aluminium chloride is fixed at 0.0200 mol. Sodium hydroxide has a concentration of 1.00 mol L⁻¹.

The precipitation reaction is:

\[ \ce{AlCl3(aq) + 3NaOH(aq) -> Al(OH)3(s) + 3NaCl(aq)} \]

Aluminium hydroxide is an amphoteric hydroxide. This means it can react with both acids and bases.

In excess sodium hydroxide, the precipitate dissolves according to the net ionic equation:

\[ \ce{Al(OH)3(s) + OH^-(aq) -> [Al(OH)4]^-(aq)} \]

Assume both reactions proceed quantitatively as written, ignoring equilibrium effects.

An idealised graph of the mass of solid aluminium hydroxide against the volume of sodium hydroxide shows:

• A rising section up to 60.0 mL.
• A maximum solid mass of 1.56 g at 60.0 mL.
• A falling section after 60.0 mL.
• Zero solid remaining at 80.0 mL.

Calculate:

a) The mass of solid aluminium hydroxide remaining at 70.0 mL.

b) The amount of aluminium present as dissolved aluminate ions at 70.0 mL.

c) The mass of solid aluminium hydroxide remaining at 90.0 mL.

d) The amount of excess hydroxide ions remaining at 90.0 mL.

Explain why the usual prediction of a horizontal plateau does not apply.

Solution 5

At 70.0 mL, 0.780 g of aluminium hydroxide remains as a solid, while 0.0100 mol of aluminium is present as dissolved aluminate ions.

At 90.0 mL, no aluminium hydroxide solid remains, and 0.0100 mol of hydroxide ions is in excess.

The important detail is that excess sodium hydroxide does not simply remain unreacted. It reacts with the precipitate itself.

Step 1: Calculate the amount of sodium hydroxide required to form the precipitate.

The fixed amount of aluminium chloride is:

\[ n(\ce{AlCl3})=0.0200\text{ mol} \]

The precipitation reaction requires three moles of NaOH for every mole of AlCl₃.

Therefore:

\[ n(\ce{NaOH})_{\text{precipitation}} =3\times0.0200 \]

\[ =0.0600\text{ mol} \]

At 1.00 mol L⁻¹, this corresponds to:

\[ V=\frac{n}{c} \]

\[ =\frac{0.0600}{1.00} \]

\[ =0.0600\text{ L} \]

\[ =60.0\text{ mL} \]

This explains why the graph reaches its maximum at 60.0 mL.

Step 2: Calculate the maximum amount and mass of aluminium hydroxide.

The equation produces one mole of Al(OH)₃ per mole of AlCl₃.

\[ n(\ce{Al(OH)3})=0.0200\text{ mol} \]

Using:

\[ M(\ce{Al(OH)3})=78.0\text{ g mol}^{-1} \]

The maximum mass is:

\[ m=nM \]

\[ =0.0200\times78.0 \]

\[ =1.56\text{ g} \]

This matches the maximum shown on the graph.

Step 3: Calculate what happens at 70.0 mL.

The amount of sodium hydroxide added is:

\[ n=cV \]

\[ =1.00\times0.0700 \]

\[ =0.0700\text{ mol} \]

The first 0.0600 mol is used to produce aluminium hydroxide.

Therefore, the amount of hydroxide remaining after precipitation is:

\[ 0.0700-0.0600 \]

\[ =0.0100\text{ mol} \]

This excess hydroxide reacts with the precipitate.

The dissolution reaction has a 1:1 ratio between hydroxide ions and aluminium hydroxide.

Therefore, 0.0100 mol of Al(OH)₃ dissolves.

The original precipitate amount was 0.0200 mol.

The amount remaining as a solid is:

\[ n_{\text{solid remaining}} =0.0200-0.0100 \]

\[ =0.0100\text{ mol} \]

Convert this into mass:

\[ m=nM \]

\[ =0.0100\times78.0 \]

\[ \boxed{0.780\text{ g}} \]

The dissolved aluminate amount is:

\[ \boxed{0.0100\text{ mol}} \]

The aluminium has not disappeared. Half remains in the solid precipitate, and half is now present in the dissolved aluminate ions.

Step 4: Calculate what happens at 90.0 mL.

The total amount of sodium hydroxide added is:

\[ n=cV \]

\[ =1.00\times0.0900 \]

\[ =0.0900\text{ mol} \]

The precipitation reaction consumes 0.0600 mol.

The remaining amount is:

\[ 0.0900-0.0600 \]

\[ =0.0300\text{ mol} \]

This additional hydroxide dissolves the aluminium hydroxide precipitate.

The dissolution reaction requires one mole of hydroxide per mole of Al(OH)₃.

Therefore, dissolving the entire 0.0200 mol of precipitate consumes:

\[ 0.0200\text{ mol OH}^- \]

The amount of hydroxide remaining is:

\[ 0.0300-0.0200 \]

\[ \boxed{0.0100\text{ mol}} \]

All the aluminium hydroxide has dissolved, so:

\[ \boxed{m(\ce{Al(OH)3})=0\text{ g}} \]

All the aluminium is now present in the dissolved aluminate form.

Step 5: Explain the unusual graph shape.

The usual product-mass graph assumes that once the fixed reactant is consumed, the excess reactant cannot consume the product.

That assumption works for our magnesium hydroxide example under the stated conditions.

It does not work here.

The first reaction produces solid aluminium hydroxide. The second reaction consumes that solid when excess hydroxide is added.

Consequently, the graph:

1. Rises while aluminium hydroxide is forming.
2. Reaches a maximum when all the aluminium chloride has reacted.
3. Falls as excess hydroxide dissolves the precipitate.

This is an important limitation of the standard limiting-reagent graph method. Always check whether excess reactant can react with the product before assuming the graph must become horizontal.

For ordinary HSC limiting-reagent graphs, remember the central decision: locate the breakpoint, compare the available moles divided by their coefficients, and use the limiting reagent to calculate theoretical yield and leftovers.

Once you can do that confidently, the next useful skill is distinguishing theoretical yield from actual yield, particularly when an experimental product-mass graph does not reach its predicted maximum.