How to Solve Gas Stoichiometry Problems at Nonstandard Conditions

A sample of magnesium reacts with excess hydrochloric acid, producing hydrogen gas. You calculate the amount of hydrogen using the balanced equation, but what volume will that hydrogen occupy if the temperature is 23°C and the pressure is 98 kPa?

You cannot use a fixed molar gas volume because the conditions are not necessarily standard. Instead, calculate the amount of gas using stoichiometry, then use the ideal gas equation, \(PV=nRT\), to find its volume at the actual temperature and pressure.

The key sequence is:

1. Write and balance the chemical equation.
2. Calculate the amount of the limiting reactant.
3. Use the mole ratio to calculate the amount of gas produced.
4. Convert the temperature to kelvin and check the pressure.
5. Use \(PV=nRT\) to calculate the gas volume.

If the gas was collected over water, you also need to correct for water vapour pressure before using the equation.

Let’s work through why this method works, how experimental measurements affect the calculation, and how to handle more demanding HSC-style questions.

01Why gas stoichiometry changes at nonstandard conditions

Consider the reaction between magnesium and hydrochloric acid:

\[
\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)}
\]

The balanced equation tells us that one mole of magnesium produces one mole of hydrogen, provided hydrochloric acid is available in sufficient quantity.

Suppose you react \(0.250\ \mathrm{g}\) of magnesium.

You can calculate the amount of hydrogen produced from the mass of magnesium. However, the balanced equation does not tell you directly how many litres of hydrogen will form.

Why not?

Because the same amount of gas can occupy different volumes depending on its temperature and pressure.

• Increasing temperature generally increases the volume of a gas at constant pressure.
• Increasing pressure decreases the volume of a gas at constant temperature.

This is why a gas stoichiometry problem has two distinct stages:

• Chemical stoichiometry determines how many moles of gas are produced.
• The gas equation determines what volume those moles occupy under the specified conditions.

Do not confuse these stages. The mole ratio does not change when temperature or pressure changes. The gas volume does.

02The experiment: collecting a gas over water

A common way to investigate gas stoichiometry is to react a solid with an acid and collect the gas produced.

Consider an experiment involving magnesium and hydrochloric acid.

The apparatus is arranged as follows:

1. A known mass of magnesium is placed inside a conical flask.
2. Dilute hydrochloric acid is held in a dropping funnel connected through the flask’s stopper.
3. Opening the funnel starts the reaction without opening the apparatus.
4. Hydrogen travels from the flask through a delivery tube.
5. The tube carries the gas into an inverted, water-filled graduated measuring cylinder.
6. Hydrogen collects at the top of the cylinder, displacing water.
7. The volume is read once the gas has reached the temperature of its surroundings.

A thermometer measures the temperature, while a barometer provides atmospheric pressure.

What should the experiment show?

The balanced equation predicts that one mole of magnesium produces one mole of hydrogen.

If all the magnesium reacts, we can calculate the theoretical amount of hydrogen.

However, the measuring cylinder does not contain perfectly dry hydrogen. It contains hydrogen mixed with water vapour.

This matters because the measured pressure is the combined pressure of both gases.

There is another experimental detail. The water levels inside and outside the measuring cylinder must be equal before we can assume that the total gas pressure equals atmospheric pressure.

An initial calculation that ignores these details can produce an incorrect amount of hydrogen, even if the chemical equation and mole ratio are completely correct.

The revised method therefore requires three things:

• Allow the gas to reach thermal equilibrium with its surroundings.
• Equalise the water levels before reading the volume, or correct for their difference.
• Correct the measured pressure for water vapour.

These are corrections to the experimental measurements. They do not change the chemical equation or the value of the gas constant.

03The ideal gas equation and its units

Once you know the amount of gas and its actual conditions, use the ideal gas equation:

\[
PV=nRT
\]

The symbols represent:

If you are calculating gas volume, rearrange the equation:

\[
V=\frac{nRT}{P}
\]

If you are calculating the amount of gas from an experimental volume:

\[
n=\frac{PV}{RT}
\]

The correct value of \(R\) depends on the pressure and volume units you use.

Choosing the correct value of R

For most calculations using pressure in kilopascals and volume in litres, use:

\[
R=8.314\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}
\]

Other equivalent forms include:

Always include the appropriate units.

Notice that \(8.314\) works for both kPa and L and Pa and cubic metres. This is because:

\[
1\ \mathrm{kPa\,L}=1\ \mathrm J
\]

If you want a closer look at why these values are equivalent, see Choosing R When Pressure Is in kPa and Volume Is in L.

Temperature must be in kelvin

The ideal gas equation uses absolute temperature, measured in kelvin.

Convert from degrees Celsius using:

\[
T(\mathrm K)=t(^{\circ}\mathrm C)+273.15
\]

For example:

\[
23.0^\circ\mathrm C=296.15\ \mathrm K
\]

Never substitute \(23.0\) directly into the equation.

The Celsius scale starts at the freezing point of water, whereas the kelvin scale starts at absolute zero. Gas-law relationships involving temperature require the absolute scale.

Pressure must be absolute

This is another important detail.

Absolute pressure is measured relative to a perfect vacuum. Gauge pressure is measured relative to the surrounding atmospheric pressure.

If a question gives gauge pressure, convert it using:

\[
P_{\mathrm{absolute}}=
P_{\mathrm{gauge}}+P_{\mathrm{atmospheric}}
\]

For example, suppose a pressure gauge reads \(25.0\ \mathrm{kPa}\), and atmospheric pressure is \(100.0\ \mathrm{kPa}\).

Then:

\[
P_{\mathrm{absolute}}=25.0+100.0
\]

\[
P_{\mathrm{absolute}}=125.0\ \mathrm{kPa}
\]

Use \(125.0\ \mathrm{kPa}\), not \(25.0\ \mathrm{kPa}\), in the ideal gas equation.

04Correcting experimental gas measurements

There is an important difference between correcting an experimental measurement and correcting for non-ideal gas behaviour.

Let’s look at the main situations you might encounter.

1. Gas collected over water

When a gas is collected over water, some water evaporates into the collection cylinder.

The cylinder therefore contains a mixture of the desired gas and water vapour.

According to Dalton’s law of partial pressures:

\[
P_{\mathrm{total}}=
P_{\mathrm{gas}}+P_{\ce{H2O}}
\]

Rearranging:

\[
P_{\mathrm{gas}}=
P_{\mathrm{total}}-P_{\ce{H2O}}
\]

If the water levels inside and outside the cylinder are equal, the total pressure equals atmospheric pressure.

Therefore:

\[
P_{\mathrm{gas}}=
P_{\mathrm{atmospheric}}-P_{\ce{H2O}}
\]

For example, suppose:

• Atmospheric pressure is \(101.3\ \mathrm{kPa}\).
• Water vapour pressure is \(3.17\ \mathrm{kPa}\).

The pressure of the dry gas is:

\[
P_{\mathrm{gas}}=101.3-3.17
\]

\[
P_{\mathrm{gas}}=98.13\ \mathrm{kPa}
\]

You must use \(98.13\ \mathrm{kPa}\) in your gas-law calculation.

One limitation is worth remembering: subtracting water vapour pressure corrects the pressure contribution of water vapour. It does not correct for gas escaping or dissolving in the water. This is particularly relevant when collecting gases such as carbon dioxide.

2. Water levels are not equal

If the water level inside the collection cylinder differs from the level outside, the gas pressure is not necessarily atmospheric pressure.

A difference in water height creates a hydrostatic pressure difference.

For a typical HSC calculation, the question may provide the corresponding pressure correction. Apply it before subtracting water vapour pressure.

Always pay attention to which water level is higher.

If the water level inside the inverted cylinder is higher than the outside level, the gas pressure is lower than atmospheric pressure.

3. Real gases versus ideal gases

The ideal gas equation is a model. It assumes:

• Gas particles occupy negligible volume compared with the container.
• Intermolecular attractions and repulsions are negligible.
• Collisions between particles are elastic.

Real gases do not satisfy these assumptions perfectly.

At relatively high pressures or low temperatures, intermolecular forces and the volume of gas particles can become significant.

The ideal gas equation may then become less accurate.

This is different from the experimental corrections above.

The key distinction is that experimental corrections improve the values substituted into the equation. Non-ideal behaviour is a limitation of the equation’s assumptions.

For an advanced extension, a real-gas correction can be represented using the compressibility factor \(Z\):

\[
PV=ZnRT
\]

An ideal gas has \(Z=1\). Only introduce this correction when the question specifically supplies or requests it.

05A reliable method for solving gas stoichiometry problems

Before working through examples, establish the order of your calculations.

There are two common directions for these questions.

If you are predicting a gas volume from a reactant mass:

\[
\text{Mass}\rightarrow
\text{Moles of reactant}\rightarrow
\text{Moles of gas}\rightarrow
\text{Gas volume}
\]

If you are using an experimental gas volume to find a reactant amount:

\[
\text{Corrected gas data}\rightarrow
\text{Moles of gas}\rightarrow
\text{Moles of reactant}\rightarrow
\text{Reactant mass}
\]

The chemistry and gas-law calculations are the same. You are simply working in opposite directions.

06Worked examples

07Questions and solutions

The following questions move from standard gas-law calculations towards more demanding experimental and conceptual problems. Treat the questions as independent scenarios unless stated otherwise.

Question 1

A \(1.68\ \mathrm g\) sample of sodium hydrogen carbonate reacts completely with excess hydrochloric acid.

The carbon dioxide produced is collected as a dry gas at \(27.0^\circ\mathrm C\) and an absolute pressure of \(98.0\ \mathrm{kPa}\).

The reaction is:

\[
\ce{NaHCO3 + HCl -> NaCl + H2O + CO2}
\]

Calculate the volume of carbon dioxide produced.

Solution 1

The volume of carbon dioxide produced is \(0.509\ \mathrm L\).

The first step is to calculate the amount of sodium hydrogen carbonate.

Its molar mass is \(84.01\ \mathrm{g\,mol^{-1}}\).

\[
n(\ce{NaHCO3})=\frac{1.68}{84.01}
\]

\[
n(\ce{NaHCO3})=0.0200\ \mathrm{mol}
\]

The balanced equation gives a 1:1 mole ratio between sodium hydrogen carbonate and carbon dioxide.

Therefore:

\[
n(\ce{CO2})=0.0200\ \mathrm{mol}
\]

Convert the temperature:

\[
T=27.0+273.15=300.15\ \mathrm K
\]

The gas is dry, so no water vapour correction is required.

Now use:

\[
V=\frac{nRT}{P}
\]

\[
V=\frac{(0.0200)(8.314)(300.15)}
{98.0}
\]

\[
V=0.509\ \mathrm L
\]

The reaction produces \(0.509\ \mathrm L\) of carbon dioxide under the specified conditions.

The important point is that the mole ratio determines the amount of carbon dioxide, while the temperature and pressure determine its volume.

Question 2

A \(2.50\ \mathrm g\) sample of zinc reacts with \(80.0\ \mathrm{mL}\) of \(1.00\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.

Hydrogen is collected as a dry gas in a gas syringe at \(40.0^\circ\mathrm C\) and an absolute pressure of \(96.0\ \mathrm{kPa}\).

The reaction is:

\[
\ce{Zn + 2HCl -> ZnCl2 + H2}
\]

Calculate the theoretical volume of hydrogen produced.

Solution 2

The theoretical volume of hydrogen produced is \(1.04\ \mathrm L\).

The key is to identify the limiting reagent before using the gas equation.

Step 1: Calculate the amount of zinc.

The molar mass of zinc is \(65.38\ \mathrm{g\,mol^{-1}}\).

\[
n(\ce{Zn})=\frac{2.50}{65.38}
\]

\[
n(\ce{Zn})=0.0382\ \mathrm{mol}
\]

Step 2: Calculate the amount of hydrochloric acid.

Convert the volume:

\[
80.0\ \mathrm{mL}=0.0800\ \mathrm L
\]

Then:

\[
n=cV
\]

\[
n(\ce{HCl})=1.00\times0.0800
\]

\[
n(\ce{HCl})=0.0800\ \mathrm{mol}
\]

Step 3: Identify the limiting reagent.

The reaction requires two moles of HCl for every mole of zinc.

The amount of HCl required to react with all the zinc is:

\[
n(\ce{HCl})=2(0.0382)
\]

\[
n(\ce{HCl})=0.0765\ \mathrm{mol}
\]

We have \(0.0800\ \mathrm{mol}\) available.

Therefore, hydrochloric acid is in excess, and zinc is the limiting reagent.

Step 4: Calculate the amount of hydrogen.

The balanced equation gives a 1:1 ratio between zinc and hydrogen.

\[
n(\ce{H2})=0.0382\ \mathrm{mol}
\]

Step 5: Convert the temperature.

\[
T=40.0+273.15
\]

\[
T=313.15\ \mathrm K
\]

Step 6: Calculate the volume.

\[
V=\frac{nRT}{P}
\]

\[
V=\frac{(0.03824)(8.314)(313.15)}
{96.0}
\]

\[
V=1.04\ \mathrm L
\]

The theoretical hydrogen volume is \(1.04\ \mathrm L\).

A tempting mistake is to assume the acid is limiting because its amount is smaller than the amount of zinc. That comparison is meaningless without considering the coefficients in the balanced equation.

Question 3

A \(0.300\ \mathrm g\) sample of magnesium contains an inert impurity that does not react with hydrochloric acid.

The sample reacts with excess hydrochloric acid, producing hydrogen collected over water.

The following illustrative experimental measurements are recorded:

The water levels inside and outside the measuring cylinder are equal.

Calculate the percentage purity of the magnesium sample.

Solution 3

The magnesium sample is approximately \(87.2\%\) pure.

The important step is to calculate the amount of dry hydrogen collected, rather than treating the entire gas mixture as hydrogen.

Step 1: Write the balanced equation.

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

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

Step 2: Correct the pressure.

The measured gas contains water vapour.

Therefore:

\[
P_{\ce{H2}}=100.2-3.17
\]

\[
P_{\ce{H2}}=97.03\ \mathrm{kPa}
\]

Step 3: Convert the volume and temperature.

\[
V=275\ \mathrm{mL}=0.275\ \mathrm L
\]

\[
T=25.0+273.15
\]

\[
T=298.15\ \mathrm K
\]

Step 4: Calculate the amount of hydrogen.

Use:

\[
n=\frac{PV}{RT}
\]

Substitute:

\[
n(\ce{H2})=
\frac{(97.03)(0.275)}
{(8.314)(298.15)}
\]

\[
n(\ce{H2})=0.01076\ \mathrm{mol}
\]

Step 5: Calculate the mass of magnesium that reacted.

The mole ratio is 1:1.

\[
n(\ce{Mg})=0.01076\ \mathrm{mol}
\]

Using the molar mass of magnesium:

\[
m=nM
\]

\[
m(\ce{Mg})=0.01076\times24.305
\]

\[
m(\ce{Mg})=0.2616\ \mathrm g
\]

Step 6: Calculate the percentage purity.

\[
\text{Purity}=
\frac{\text{Mass of pure Mg}}
{\text{Total sample mass}}\times100
\]

\[
=\frac{0.2616}{0.300}\times100
\]

\[
=87.2\%
\]

The magnesium sample is \(87.2\%\) pure.

The tempting mistake is to use \(100.2\ \mathrm{kPa}\) directly as the hydrogen pressure. This would incorrectly include the pressure contributed by water vapour, overestimating the amount of magnesium that reacted.

This calculation also assumes negligible gas loss and that the inert impurity does not produce hydrogen.

Question 4

A student collects hydrogen from the reaction between magnesium and excess hydrochloric acid using an inverted measuring cylinder.

The following measurements are recorded:

• Atmospheric pressure: \(100.0\ \mathrm{kPa}\).
• Gas volume: \(300\ \mathrm{mL}\).
• Temperature: \(20.0^\circ\mathrm C\).
• Water vapour pressure: \(2.34\ \mathrm{kPa}\).

However, the water level inside the measuring cylinder is \(8.0\ \mathrm{cm}\) higher than the water level outside it.

The resulting hydrostatic pressure difference means that the pressure of the gas mixture inside the cylinder is \(0.78\ \mathrm{kPa}\) lower than atmospheric pressure.

Calculate the mass of magnesium that reacted.

Explain why subtracting only the water vapour pressure from atmospheric pressure would produce an incorrect answer.

Solution 4

The mass of magnesium that reacted is approximately \(0.290\ \mathrm g\).

There are two separate pressure corrections. Both must be applied before using the ideal gas equation.

Step 1: Correct for the unequal water levels.

The internal water level is higher than the external level.

This means the gas pressure is lower than atmospheric pressure.

Therefore:

\[
P_{\mathrm{total}}=100.0-0.78
\]

\[
P_{\mathrm{total}}=99.22\ \mathrm{kPa}
\]

Step 2: Correct for water vapour pressure.

The gas mixture contains hydrogen and water vapour.

Subtract the water vapour pressure:

\[
P_{\ce{H2}}=99.22-2.34
\]

\[
P_{\ce{H2}}=96.88\ \mathrm{kPa}
\]

Notice that we did not subtract water vapour pressure directly from atmospheric pressure. The hydrostatic correction had to be applied first.

Step 3: Convert the temperature and volume.

\[
T=20.0+273.15=293.15\ \mathrm K
\]

\[
V=300\ \mathrm{mL}=0.300\ \mathrm L
\]

Step 4: Calculate the amount of hydrogen.

\[
n=\frac{PV}{RT}
\]

\[
n(\ce{H2})=
\frac{(96.88)(0.300)}
{(8.314)(293.15)}
\]

\[
n(\ce{H2})=0.0119\ \mathrm{mol}
\]

Step 5: Use the mole ratio to find the amount of magnesium.

The balanced equation is:

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

The ratio is 1:1.

Therefore:

\[
n(\ce{Mg})=0.0119\ \mathrm{mol}
\]

Step 6: Calculate the mass of magnesium.

\[
m=nM
\]

\[
m(\ce{Mg})=0.0119\times24.305
\]

\[
m(\ce{Mg})=0.290\ \mathrm g
\]

The mass of magnesium that reacted is \(0.290\ \mathrm g\).

The conceptual mistake is assuming that the gas pressure must equal atmospheric pressure whenever a gas is collected over water.

That is only true when the water levels are equal. A difference in water height creates an additional pressure difference, which must be accounted for independently of water vapour pressure.

Question 5

A chemistry student receives a \(4.00\ \mathrm g\) solid mixture containing three substances:

• Sodium carbonate, \(\ce{Na2CO3}\).
• Sodium hydrogen carbonate, \(\ce{NaHCO3}\).
• Sodium chloride, \(\ce{NaCl}\).

The student adds \(0.0500\ \mathrm{mol}\) of hydrochloric acid, which is known to be in excess.

Both carbonate substances react completely, producing carbon dioxide. Sodium chloride does not react.

The carbon dioxide is collected as a dry gas in a gas syringe.

The illustrative experimental data are:

Use the following molar masses:

• \(\ce{Na2CO3}\): \(106.0\ \mathrm{g\,mol^{-1}}\).
• \(\ce{NaHCO3}\): \(84.0\ \mathrm{g\,mol^{-1}}\).
• \(\ce{NaCl}\): \(58.44\ \mathrm{g\,mol^{-1}}\).

A student claims that the original mass of the mixture and the measured carbon dioxide volume are sufficient to determine the amount of all three substances.

Is the student’s claim correct?

Demonstrate your reasoning by constructing two different possible mixture compositions that satisfy the experimental measurements. Explain what additional experimental measurement would be needed to determine the composition uniquely.

Solution 5

No. The student’s claim is incorrect. The available measurements do not provide enough independent information to determine all three amounts uniquely.

The subtle issue is that the two carbonate substances both produce one mole of carbon dioxide per mole of solid, but they require different amounts of hydrochloric acid.

Step 1: Calculate the amount of carbon dioxide.

First convert the temperature:

\[
T=30.0+273.15
\]

\[
T=303.15\ \mathrm K
\]

Use the ideal gas equation:

\[
n=\frac{PV}{RT}
\]

\[
n(\ce{CO2})=
\frac{(100.0)(0.499)}
{(8.314)(303.15)}
\]

\[
n(\ce{CO2})=0.0198\ \mathrm{mol}
\]

Step 2: Write the two reaction equations.

Sodium carbonate reacts with hydrochloric acid:

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

Sodium hydrogen carbonate reacts with hydrochloric acid:

\[
\ce{NaHCO3 + HCl -> NaCl + H2O + CO2}
\]

Both produce one mole of carbon dioxide per mole of carbonate-containing solid.

However, their hydrochloric acid requirements differ:

• One mole of sodium carbonate requires two moles of HCl.
• One mole of sodium hydrogen carbonate requires one mole of HCl.

Step 3: Define the unknown amounts.

Let:

• \(x\) = moles of sodium carbonate.
• \(y\) = moles of sodium hydrogen carbonate.
• \(z\) = moles of sodium chloride.

Because both carbonate substances produce one mole of carbon dioxide per mole:

\[
x+y=0.0198
\]

The total mass gives another equation:

\[
106.0x+84.0y+58.44z=4.00
\]

We now have two independent equations but three unknowns.

There is not enough information to calculate a unique value for every variable.

Step 4: Demonstrate two possible compositions.

Consider the following first possibility:

\[
x=0.00500\ \mathrm{mol}
\]

\[
y=0.0148\ \mathrm{mol}
\]

Together, they produce:

\[
0.00500+0.0148=0.0198\ \mathrm{mol}
\]

of carbon dioxide.

Their combined mass is:

\[
m=(0.00500)(106.0)+(0.0148)(84.0)
\]

\[
m=0.530+1.2432
\]

\[
m=1.7732\ \mathrm g
\]

The remaining mass must be sodium chloride:

\[
m(\ce{NaCl})=4.00-1.7732
\]

\[
m(\ce{NaCl})=2.2268\ \mathrm g
\]

This corresponds to approximately \(0.0381\ \mathrm{mol}\) of sodium chloride.

Now consider a different composition:

\[
x=0.0148\ \mathrm{mol}
\]

\[
y=0.00500\ \mathrm{mol}
\]

Again:

\[
x+y=0.0198\ \mathrm{mol}
\]

The combined mass of the carbonate substances is now:

\[
m=(0.0148)(106.0)+(0.00500)(84.0)
\]

\[
m=1.5688+0.420
\]

\[
m=1.9888\ \mathrm g
\]

The remaining sodium chloride mass is:

\[
m(\ce{NaCl})=4.00-1.9888
\]

\[
m(\ce{NaCl})=2.0112\ \mathrm g
\]

This corresponds to approximately \(0.0344\ \mathrm{mol}\) of sodium chloride.

Both compositions satisfy the measured total mass and produce the same amount of carbon dioxide.

Step 5: Check whether the hydrochloric acid information distinguishes the possibilities.

For the first mixture, the amount of hydrochloric acid required is:

\[
n(\ce{HCl})=2x+y
\]

\[
=2(0.00500)+0.0148
\]

\[
=0.0248\ \mathrm{mol}
\]

For the second mixture:

\[
n(\ce{HCl})=2(0.0148)+0.00500
\]

\[
=0.0346\ \mathrm{mol}
\]

Both amounts are less than the \(0.0500\ \mathrm{mol}\) supplied.

Therefore, both mixtures are consistent with hydrochloric acid being in excess.

Knowing that the acid is in excess gives us an inequality, not another exact equation. It does not distinguish between the two possible compositions.

Step 6: Identify the additional measurement.

We need an independent measurement of the amount of hydrochloric acid actually consumed.

For example, we could measure the excess hydrochloric acid remaining after the reaction using an appropriate titration.

Once the amount of acid consumed is known, we have:

\[
x+y=n(\ce{CO2})
\]

and:

\[
2x+y=n(\ce{HCl})_{\mathrm{consumed}}
\]

Subtracting the first equation from the second gives:

\[
x=n(\ce{HCl})_{\mathrm{consumed}}-n(\ce{CO2})
\]

Then:

\[
y=2n(\ce{CO2})-
n(\ce{HCl})_{\mathrm{consumed}}
\]

Finally, the original mass equation can be used to determine \(z\).

We now have enough independent information to calculate all three amounts, assuming no other substances consume hydrochloric acid.

The important lesson is that having several measurements does not automatically mean you have enough independent equations to determine every unknown.

The gas-law calculation converts between the amount of gas and its volume at specified conditions. The chemical equation connects that amount back to the reacting substances.

Once those two stages are reliable, the next decision is whether an experimental result differs from the theoretical prediction because of a limiting reagent, incomplete reaction, sample impurity, or gas loss. Those are different problems, and separating them is essential when interpreting quantitative chemistry experiments.