How to Find the Reacted Amount in Molar Enthalpy Calculations

Suppose you place \(0.0200\ \mathrm{mol}\) of magnesium into a solution containing \(0.0300\ \mathrm{mol}\) of hydrochloric acid. The balanced equation is:

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

You started with \(0.0200\) mol of magnesium. Does that mean \(0.0200\) mol reacted?

No. Each mole of magnesium needs two moles of hydrochloric acid. The available acid can react with only \(0.0150\) mol of magnesium, leaving \(0.00500\) mol unreacted.

If you’re calculating the molar enthalpy per mole of magnesium consumed, your denominator must therefore be \(0.0150\ \mathrm{mol}\), not \(0.0200\ \mathrm{mol}\).

The rule is simple: calculate the amount that actually reacted, not merely the amount you started with. Use the balanced equation and limiting reactant if the reaction goes to completion. If it stops early, use experimental evidence to determine how much actually reacted.

01First separate the heat from the amount reacted

There are two different calculations involved in finding molar enthalpy:

• Calculate how much heat the reaction transferred.
• Calculate how many moles of the relevant substance actually reacted.

These calculations have different purposes. Mixing them up is a common source of incorrect answers.

System and surroundings: where does the heat go?

Imagine adding magnesium to hydrochloric acid inside an insulated cup.

The reacting chemicals are the system. The system is where the chemical change takes place.

The surroundings are everything receiving or supplying heat during the experiment. In a typical school calorimetry experiment, these include the solution, the calorimeter, and the thermometer.

When magnesium reacts with hydrochloric acid, the reaction releases energy. The surrounding solution absorbs that energy, so its temperature increases.

The important point is that the thermometer measures the temperature change of the surroundings, not the heat released directly by the reaction.

For an insulated system, conservation of energy gives:

\[ q_{\mathrm{system}}+q_{\mathrm{surroundings}}=0 \]

Therefore:

\[ \boxed{q_{\mathrm{reaction}}=-q_{\mathrm{surroundings}}} \]

The negative sign matters. It tells us that the reaction and its surroundings exchange heat in opposite directions.

Understanding the sign of \(q\)

The symbol \(q\) represents heat transferred.

Its sign depends on whether the substance being considered gains or loses heat.

For an exothermic reaction, the system releases heat, so its enthalpy change is negative.

For an endothermic reaction, the system absorbs heat, so its enthalpy change is positive.

Don’t automatically give the reaction a positive enthalpy change because the thermometer shows a positive temperature change. The thermometer measures the surroundings.

Calculating the heat transferred

For a solution calorimetry experiment, the heat absorbed by the surroundings is commonly calculated using:

\[ q=mc\Delta T \]

where:

• \(q\) is heat transferred, in joules (J).
• \(m\) is the mass of the solution, in grams (g).
• \(c\) is the specific heat capacity, in \(\mathrm{J\,g^{-1}\,^{\circ}C^{-1}}\).
• \(\Delta T\) is the temperature change, in °C.

Calculate the temperature change using:

\[ \Delta T=T_{\mathrm{final}}-T_{\mathrm{initial}} \]

For dilute aqueous solutions, questions commonly assume:

\[ c=4.18\ \mathrm{J\,g^{-1}\,^{\circ}C^{-1}} \]

If the calorimeter has a supplied heat capacity, include its contribution as well:

\[ q_{\mathrm{surroundings}} =mc\Delta T+C_{\mathrm{cal}}\Delta T \]

Here, \(C_{\mathrm{cal}}\) is the calorimeter’s heat capacity in \(\mathrm{J\,^{\circ}C^{-1}}\).

If the question says the calorimeter absorbs negligible heat, you can ignore this additional term.

At approximately constant atmospheric pressure, the heat transferred by the reaction corresponds to its enthalpy change.

For a more detailed explanation of the temperature calculation, see How to Calculate Enthalpy Change from a Temperature Rise.

02Finding the amount that actually reacted

Now we can deal with the main question: what number belongs in the denominator when calculating molar enthalpy?

The answer depends on the balanced equation and the quantity of substance the question asks you to report the enthalpy against.

Step 1: Calculate the initial moles of each reactant

Start with the quantities supplied in the question.

If you have a solution with a known concentration and volume, use:

\[ n=CV \]

where:

• \(n\) is the amount in moles.
• \(C\) is the concentration in \(\mathrm{mol\,L^{-1}}\).
• \(V\) is the volume in litres.

If you are given a mass instead, use:

\[ n=\frac{m}{M_{\mathrm{m}}} \]

where:

• \(m\) is the mass in grams.
• \(M_{\mathrm{m}}\) is the molar mass in \(\mathrm{g\,mol^{-1}}\).

Remember to convert millilitres into litres before using \(n=CV\).

If you need to revise mass-to-mole conversions, see Moles from Mass and Molar Mass: HSC Chemistry Guide.

Step 2: Use the balanced equation to identify the limiting reactant

The limiting reactant is the reactant that runs out first when a reaction proceeds to completion.

It determines the maximum amount of reaction that can occur.

Consider:

\[ \ce{2H2 + O2 -> 2H2O} \]

The equation tells us that two moles of hydrogen react with one mole of oxygen.

Suppose you have:

• \(0.0400\) mol of hydrogen.
• \(0.0300\) mol of oxygen.

You might initially think oxygen is limiting because it has fewer moles.

But compare the amounts against their stoichiometric coefficients:

Hydrogen is limiting because it runs out first.

Only \(0.0200\) mol of reaction, as represented by the balanced equation, can occur.

This is called the extent of reaction, represented by \(\xi\).

For a general reaction:

\[ \ce{aA + bB -> cC + dD} \]

the maximum extent of reaction is:

\[ \xi_{\max}= \min\left( \frac{n(A)_0}{a}, \frac{n(B)_0}{b} \right) \]

Here, \(n(A)_0\) and \(n(B)_0\) are the initial amounts of the reactants.

The actual amounts consumed or formed are then:

\[ n(A)_{\mathrm{reacted}}=a\xi \]

\[ n(B)_{\mathrm{reacted}}=b\xi \]

\[ n(C)_{\mathrm{formed}}=c\xi \]

This distinction is important: the amount of the limiting reactant is not necessarily equal to the extent of reaction. You must account for its coefficient.

Step 3: Choose the correct amount for the molar enthalpy

This is where you need to read the question carefully.

Molar enthalpy is an enthalpy change expressed per mole of a specified chemical amount. That amount depends on how the reaction is being described.

For example, consider:

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

If \(0.0300\) mol of HCl reacts completely, then:

• \(0.0150\) mol of Mg reacts.
• \(0.0300\) mol of HCl reacts.
• \(0.0150\) mol of \(\ce{MgCl2}\) forms.
• \(0.0150\) mol of \(\ce{H2}\) forms.

These are different amounts describing the same reaction.

You can express the enthalpy per mole of magnesium consumed or per mole of hydrochloric acid consumed. The numerical values will differ because the denominators differ.

The key is to use the amount that matches the wording of the question.

What if the reaction doesn’t go to completion?

There is one more complication.

The limiting reactant tells you how much reaction could occur if the reaction proceeds to completion. It does not always tell you how much actually occurred.

Imagine adding magnesium to excess hydrochloric acid but stopping the experiment before all the magnesium reacts.

The limiting reagent calculation predicts a maximum of \(0.0100\) mol of magnesium could react.

However, suppose a gas syringe shows that only \(0.00800\) mol of hydrogen was produced.

From the balanced equation:

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

one mole of magnesium produces one mole of hydrogen.

Therefore, only \(0.00800\) mol of magnesium actually reacted.

In this situation, the measured product amount is more useful than the theoretical maximum.

For standard HSC-style questions, a reaction is generally treated as complete when the question gives no indication otherwise. Always follow any additional experimental information provided.

Important measurement assumptions

Before calculating molar enthalpy, check what the question allows you to assume.

These are assumptions, not universal facts.

For example, a spirit burner can lose fuel through evaporation without combustion. If an unlit control experiment measures evaporation, subtract that loss before calculating the moles of fuel actually burnt.

Also remember that the mass used in \(q=mc\Delta T\) and the amount used in the molar enthalpy calculation have different jobs.

The mass represents the surroundings absorbing heat. The moles represent the chemical change responsible for that heat.

03Worked examples

All numerical data in the following examples are illustrative.

04Questions and solutions

The following questions build from straightforward limiting-reactant calculations to situations where the theoretical amount, measured amount, and total heat released need to be considered separately.

Question 1

A student mixes \(40.0\ \mathrm{mL}\) of \(0.500\ \mathrm{mol\,L^{-1}}\) hydrochloric acid with \(30.0\ \mathrm{mL}\) of \(0.500\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.

The temperature increases from \(19.8\,^{\circ}\mathrm{C}\) to \(22.7\,^{\circ}\mathrm{C}\).

Assume:

• Both solutions have a density of \(1.00\ \mathrm{g\,mL^{-1}}\).
• The specific heat capacity is \(4.18\ \mathrm{J\,g^{-1}\,^{\circ}C^{-1}}\).
• The calorimeter absorbs negligible heat.
• The reaction goes to completion.

Calculate the molar enthalpy of neutralisation per mole of water formed.

Solution 1

The molar enthalpy of neutralisation is \(-56.6\ \mathrm{kJ\,mol^{-1}}\).

First, calculate the initial moles of each reactant.

For hydrochloric acid:

\[ n(\ce{HCl})=0.500\times0.0400 =0.0200\ \mathrm{mol} \]

For sodium hydroxide:

\[ n(\ce{NaOH})=0.500\times0.0300 =0.0150\ \mathrm{mol} \]

The balanced equation is:

\[ \ce{HCl + NaOH -> NaCl + H2O} \]

The mole ratio is 1:1, so sodium hydroxide is limiting.

The amount of water formed is:

\[ n(\ce{H2O})=0.0150\ \mathrm{mol} \]

Next, calculate the total mass of the solution:

\[ m=40.0+30.0=70.0\ \mathrm{g} \]

The temperature change is:

\[ \Delta T=22.7-19.8=2.9\,^{\circ}\mathrm{C} \]

Calculate the heat absorbed:

\[ \begin{aligned} q_{\mathrm{surroundings}} &=mc\Delta T\\ &=70.0\times4.18\times2.9\\ &=848.54\ \mathrm{J}\\ &=0.849\ \mathrm{kJ} \end{aligned} \]

Therefore:

\[ q_{\mathrm{reaction}}=-0.849\ \mathrm{kJ} \]

Divide by the amount of water formed:

\[ \begin{aligned} \Delta H_{\mathrm{neut}} &=\frac{-0.84854}{0.0150}\\ &=-56.6\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The result is negative because neutralisation releases heat.

The excess hydrochloric acid still contributes to the mass of the solution, even though it does not all react.

Question 2

A student performs a combustion experiment using propan-1-ol.

The spirit burner’s mass decreases from \(75.860\) g to \(75.410\) g during the experiment.

A matched unlit control burner loses \(0.030\) g through evaporation over the same period.

The experiment also provides the following measurements:

• Mass of water: \(200.0\) g.
• Initial water temperature: \(18.5\,^{\circ}\mathrm{C}\).
• Final water temperature: \(34.7\,^{\circ}\mathrm{C}\).
• Calorimeter heat capacity: \(35.0\ \mathrm{J\,^{\circ}C^{-1}}\).
• Molar mass of propan-1-ol: \(60.0\ \mathrm{g\,mol^{-1}}\).

Assume the evaporation control accurately estimates fuel evaporation during the combustion experiment, and all corrected fuel loss undergoes complete combustion.

Calculate the experimental molar enthalpy of combustion of propan-1-ol.

Explain why using the uncorrected burner mass loss would produce an incorrect answer.

Solution 2

The experimental molar enthalpy of combustion is approximately \(-2.02\times10^3\ \mathrm{kJ\,mol^{-1}}\).

The first task is to calculate how much fuel actually combusted.

The total mass loss from the burner is:

\[ \begin{aligned} m_{\mathrm{loss}} &=75.860-75.410\\ &=0.450\ \mathrm{g} \end{aligned} \]

However, \(0.030\) g evaporated without combustion.

Therefore:

\[ \begin{aligned} m_{\mathrm{burnt}} &=0.450-0.030\\ &=0.420\ \mathrm{g} \end{aligned} \]

Step 1: Calculate the moles of propan-1-ol actually burnt.

\[ \begin{aligned} n&=\frac{m}{M_{\mathrm{m}}}\\ &=\frac{0.420}{60.0}\\ &=0.00700\ \mathrm{mol} \end{aligned} \]

This is the correct denominator for molar enthalpy of combustion.

Step 2: Calculate the temperature change.

\[ \Delta T=34.7-18.5 =16.2\,^{\circ}\mathrm{C} \]

Step 3: Calculate the heat absorbed by the water.

\[ \begin{aligned} q_{\mathrm{water}} &=mc\Delta T\\ &=200.0\times4.18\times16.2\\ &=13543.2\ \mathrm{J} \end{aligned} \]

Step 4: Include the heat absorbed by the calorimeter.

\[ \begin{aligned} q_{\mathrm{cal}} &=C_{\mathrm{cal}}\Delta T\\ &=35.0\times16.2\\ &=567\ \mathrm{J} \end{aligned} \]

The total heat absorbed by the surroundings is:

\[ \begin{aligned} q_{\mathrm{surroundings}} &=13543.2+567\\ &=14110.2\ \mathrm{J}\\ &=14.1\ \mathrm{kJ} \end{aligned} \]

Therefore:

\[ q_{\mathrm{combustion}}=-14.1\ \mathrm{kJ} \]

Step 5: Calculate the molar enthalpy of combustion.

\[ \begin{aligned} \Delta H_{\mathrm{c}} &=\frac{q_{\mathrm{combustion}}} {n(\mathrm{fuel})}\\ &=\frac{-14.1102}{0.00700}\\ &=-2015.7\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

To three significant figures:

\[ \boxed{\Delta H_{\mathrm{c}} =-2.02\times10^3\ \mathrm{kJ\,mol^{-1}}} \]

Why can’t we use the full burner mass loss?

If we used \(0.450\) g instead of \(0.420\) g:

\[ n=\frac{0.450}{60.0} =0.00750\ \mathrm{mol} \]

This would give:

\[ \Delta H_{\mathrm{c}} =\frac{-14.1102}{0.00750} =-1881\ \mathrm{kJ\,mol^{-1}} \]

The result is less negative because we have incorrectly counted evaporated fuel as fuel that reacted.

The burner lost mass, but not all of that mass underwent combustion.

Question 3

Sodium carbonate reacts with hydrochloric acid according to the equation:

\[ \ce{Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O} \]

A reaction mixture initially contains:

• \(0.0250\) mol of sodium carbonate.
• \(0.0400\) mol of hydrochloric acid.

The reaction proceeds to completion, and the corrected calorimetry measurement shows that the surroundings absorb \(0.576\) kJ of heat.

Calculate:

(a) The amount of sodium carbonate consumed.

(b) The amount of hydrochloric acid consumed.

(c) The amount of water formed.

(d) The molar enthalpy per mole of sodium carbonate consumed and per mole of hydrochloric acid consumed.

Explain why the two molar enthalpy values are different.

Solution 3

The reaction consumes \(0.0200\) mol of sodium carbonate and \(0.0400\) mol of hydrochloric acid, forming \(0.0200\) mol of water. The corresponding molar enthalpies are \(-28.8\) and \(-14.4\ \mathrm{kJ\,mol^{-1}}\), respectively.

Start with the balanced equation:

\[ \ce{Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O} \]

Step 1: Identify the limiting reactant.

The stoichiometric ratio is:

\[ 1\ \mathrm{mol\ Na2CO3}:2\ \mathrm{mol\ HCl} \]

Compare the initial amounts against their coefficients:

For sodium carbonate:

\[ \frac{0.0250}{1}=0.0250 \]

For hydrochloric acid:

\[ \frac{0.0400}{2}=0.0200 \]

Hydrochloric acid is limiting because it has the smaller coefficient-adjusted amount.

Therefore, the extent of reaction is:

\[ \xi=0.0200\ \mathrm{mol} \]

Step 2: Calculate the amount of each reactant consumed.

One mole of sodium carbonate reacts per mole of reaction:

\[ n(\ce{Na2CO3})_{\mathrm{reacted}} =1\times0.0200 \]

\[ \boxed{0.0200\ \mathrm{mol}} \]

Two moles of hydrochloric acid react per mole of reaction:

\[ n(\ce{HCl})_{\mathrm{reacted}} =2\times0.0200 \]

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

The remaining sodium carbonate is:

\[ 0.0250-0.0200 =0.00500\ \mathrm{mol} \]

Step 3: Calculate the amount of water formed.

The coefficient of water is one.

Therefore:

\[ n(\ce{H2O})=1\times0.0200 \]

\[ \boxed{0.0200\ \mathrm{mol}} \]

Step 4: Determine the heat released by the reaction.

The surroundings absorb \(0.576\) kJ.

Therefore:

\[ q_{\mathrm{reaction}}=-0.576\ \mathrm{kJ} \]

Step 5: Calculate the molar enthalpy per mole of sodium carbonate consumed.

\[ \begin{aligned} \Delta H_{\ce{Na2CO3}} &=\frac{-0.576}{0.0200}\\ &=-28.8\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Step 6: Calculate the molar enthalpy per mole of hydrochloric acid consumed.

\[ \begin{aligned} \Delta H_{\ce{HCl}} &=\frac{-0.576}{0.0400}\\ &=-14.4\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The two answers differ because they express the same total reaction heat per mole of different substances.

One mole of sodium carbonate requires two moles of hydrochloric acid. Consequently, the heat per mole of hydrochloric acid is half the heat per mole of sodium carbonate.

Neither result is inherently incorrect. The correct answer depends on the chemical basis specified in the question.

Question 4

Aluminium reacts with hydrochloric acid according to:

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

An experiment begins with:

• \(0.0150\) mol of aluminium.
• \(0.0500\) mol of hydrochloric acid.

The reaction is deliberately stopped early. A gas syringe collects \(0.0180\) mol of hydrogen.

The corrected calorimetry measurement shows that the surroundings absorb \(6.37\) kJ of heat.

Calculate:

(a) The actual amount of aluminium consumed.

(b) The amount of hydrochloric acid consumed.

(c) The molar enthalpy per mole of aluminium consumed.

(d) The amount of aluminium remaining.

Explain why identifying the limiting reactant alone is insufficient to determine the actual amount reacted in this experiment.

Solution 4

The experiment consumes \(0.0120\) mol of aluminium, and its experimental molar enthalpy is \(-531\ \mathrm{kJ\,mol^{-1}}\).

The key is to use the measured hydrogen production rather than assume the reaction went to completion.

Step 1: Calculate the actual extent of reaction.

The balanced equation is:

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

Three moles of hydrogen form per mole of reaction.

Therefore:

\[ \begin{aligned} \xi &=\frac{n(\ce{H2})}{3}\\ &=\frac{0.0180}{3}\\ &=0.00600\ \mathrm{mol} \end{aligned} \]

Step 2: Calculate the actual amount of aluminium consumed.

Two moles of aluminium react per mole of reaction.

\[ \begin{aligned} n(\ce{Al})_{\mathrm{reacted}} &=2\xi\\ &=2\times0.00600\\ &=0.0120\ \mathrm{mol} \end{aligned} \]

Step 3: Calculate the hydrochloric acid consumed.

Six moles of hydrochloric acid react per mole of reaction.

\[ \begin{aligned} n(\ce{HCl})_{\mathrm{reacted}} &=6\times0.00600\\ &=0.0360\ \mathrm{mol} \end{aligned} \]

Step 4: Calculate the molar enthalpy per mole of aluminium consumed.

The surroundings absorb heat, so:

\[ q_{\mathrm{reaction}}=-6.37\ \mathrm{kJ} \]

The required denominator is \(0.0120\) mol of aluminium.

\[ \begin{aligned} \Delta H_{\ce{Al}} &=\frac{-6.37}{0.0120}\\ &=-530.8\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{\Delta H_{\ce{Al}} =-531\ \mathrm{kJ\,mol^{-1}}} \]

Step 5: Calculate the remaining aluminium.

\[ \begin{aligned} n(\ce{Al})_{\mathrm{remaining}} &=0.0150-0.0120\\ &=0.00300\ \mathrm{mol} \end{aligned} \]

Why isn’t the limiting reactant calculation enough?

If the reaction went to completion, aluminium would be limiting.

The maximum extent would be:

\[ \xi_{\max} =\min\left(\frac{0.0150}{2}, \frac{0.0500}{6}\right) \]

\[ \xi_{\max}=0.00750\ \mathrm{mol} \]

This would allow \(0.0150\) mol of aluminium to react.

However, the measured hydrogen production shows that the actual extent was only \(0.00600\) mol.

The reaction was stopped early, so the theoretical maximum cannot be used as the actual reacted amount.

Question 5

A solid sample has a mass of \(0.646\) g and contains only magnesium and magnesium oxide.

The sample is added to excess hydrochloric acid. Both solids react completely according to the following equations:

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

\[ \ce{MgO + 2HCl -> MgCl2 + H2O} \]

The experiment produces \(0.0100\) mol of hydrogen gas.

The corrected calorimetry measurement shows that the surroundings absorb \(6.18\) kJ of heat from the combined reactions.

The molar enthalpy of the magnesium oxide reaction is supplied as:

\[ \Delta H_{\ce{MgO}}=-151\ \mathrm{kJ\,mol^{-1}} \]

Use the following molar masses:

• Magnesium: \(24.3\ \mathrm{g\,mol^{-1}}\).
• Magnesium oxide: \(40.3\ \mathrm{g\,mol^{-1}}\).

Calculate the experimental molar enthalpy of the magnesium reaction per mole of magnesium consumed.

Assume:

• All hydrogen comes from the reaction of magnesium with hydrochloric acid.
• Both reactions proceed to completion.
• The sample contains no other substances.
• All heat released by both reactions is included in the calorimetry measurement.

Explain why simply dividing the total heat released by the measured moles of hydrogen would give an incorrect result.

Solution 5

The experimental molar enthalpy of the magnesium reaction is \(-467\ \mathrm{kJ\,mol^{-1}}\).

This question has an important complication: two different reactions release heat in the same calorimeter.

The gas measurement identifies the amount of magnesium that reacted, but the total heat measurement includes energy released by both magnesium and magnesium oxide.

We must separate those contributions.

Step 1: Use the hydrogen produced to calculate the amount of magnesium.

From the first balanced equation:

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

The mole ratio between magnesium and hydrogen is 1:1.

Therefore:

\[ n(\ce{Mg})=n(\ce{H2}) \]

\[ \boxed{n(\ce{Mg})=0.0100\ \mathrm{mol}} \]

Step 2: Calculate the mass of magnesium in the original sample.

\[ \begin{aligned} m(\ce{Mg}) &=nM_{\mathrm{m}}\\ &=0.0100\times24.3\\ &=0.243\ \mathrm{g} \end{aligned} \]

Step 3: Determine the mass and moles of magnesium oxide.

The original sample contains only magnesium and magnesium oxide.

Therefore:

\[ m(\ce{MgO})=0.646-0.243 \]

\[ m(\ce{MgO})=0.403\ \mathrm{g} \]

Convert this mass into moles:

\[ \begin{aligned} n(\ce{MgO}) &=\frac{m}{M_{\mathrm{m}}}\\ &=\frac{0.403}{40.3}\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

We now know that the sample contained:

• \(0.0100\) mol of magnesium.
• \(0.0100\) mol of magnesium oxide.

Both react completely.

Step 4: Calculate the heat released by the magnesium oxide reaction.

The supplied molar enthalpy is:

\[ \Delta H_{\ce{MgO}} =-151\ \mathrm{kJ\,mol^{-1}} \]

Therefore:

\[ \begin{aligned} q_{\ce{MgO}} &=n(\ce{MgO})\Delta H_{\ce{MgO}}\\ &=0.0100\times(-151)\\ &=-1.51\ \mathrm{kJ} \end{aligned} \]

The magnesium oxide reaction releases \(1.51\) kJ.

Step 5: Isolate the heat released by the magnesium reaction.

The calorimeter measures both reactions together.

The total heat released is:

\[ q_{\mathrm{total}}=-6.18\ \mathrm{kJ} \]

The relationship is:

\[ q_{\mathrm{total}} =q_{\ce{Mg}}+q_{\ce{MgO}} \]

Rearrange:

\[ q_{\ce{Mg}}=q_{\mathrm{total}}-q_{\ce{MgO}} \]

Substitute:

\[ \begin{aligned} q_{\ce{Mg}} &=-6.18-(-1.51)\\ &=-4.67\ \mathrm{kJ} \end{aligned} \]

The magnesium reaction itself released \(4.67\) kJ.

Step 6: Calculate the molar enthalpy of the magnesium reaction.

We know that \(0.0100\) mol of magnesium reacted.

Therefore:

\[ \begin{aligned} \Delta H_{\ce{Mg}} &=\frac{q_{\ce{Mg}}}{n(\ce{Mg})}\\ &=\frac{-4.67}{0.0100}\\ &=-467\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The final answer is:

\[ \boxed{\Delta H_{\ce{Mg}} =-467\ \mathrm{kJ\,mol^{-1}}} \]

Why is dividing the total heat by the hydrogen amount incorrect?

The hydrogen measurement tells us how much magnesium reacted. It does not tell us how much heat came from that reaction alone.

Magnesium oxide also reacts with hydrochloric acid. Although it produces no hydrogen, it still releases energy.

If we simply divided the total heat by the magnesium amount:

\[ \frac{-6.18}{0.0100} =-618\ \mathrm{kJ\,mol^{-1}} \]

we would incorrectly attribute the magnesium oxide reaction’s heat to magnesium.

The correct method is to isolate the heat from the reaction being investigated before dividing by its reacted amount.

05What to check before finalising your answer

When calculating molar enthalpy, keep these three questions separate:

• How much heat was transferred? Calculate the heat gained or lost by the surroundings, then reverse the sign to obtain the reaction heat.
• How much substance actually reacted? Use the balanced equation, limiting reactant, and any experimental measurements of actual conversion.
• What is the enthalpy being expressed per mole of? Match the denominator to the named reactant, product, fuel, or reaction extent.

The final decision is not always simply to divide by the limiting reactant.

Once you have the correct molar enthalpy, the next step is evaluating the reliability of the experimental result. Heat loss to the environment, unaccounted calorimeter heat capacity, evaporation, and incomplete reactions can all change the measured value. Understanding those limitations is essential when comparing experimental enthalpies with accepted values.