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Learn how to read gas volume ratios from reaction coefficients when temperature and pressure are kept the same.

---

## Gas Volumes Follow Coefficients

The law of combining volumes states that reacting gases, and gaseous products, have volumes in simple whole-number ratios when measured at the same temperature and pressure. This law is also known as Gay-Lussac's law of combining gas volumes.

OpenStax Chemistry 2e explains that gas volume ratios in a reaction are given by the coefficients in the balanced equation when all gas volumes are measured at the same temperature and pressure in [Stoichiometry of Gaseous Substances, Mixtures, and Reactions](https://openstax.org/books/chemistry-2e/pages/9-3-stoichiometry-of-gaseous-substances-mixtures-and-reactions). Britannica also summarizes Gay-Lussac's finding that hydrogen and oxygen combine by volume in a $$2:1$$ ratio to form water in [Gay-Lussac's law of combining volumes](https://www.britannica.com/science/Gay-Lussacs-law-of-combining-volumes).

Visible text: OpenStax Chemistry 2e explains that gas volume ratios in a reaction are given by the coefficients in the balanced equation when all gas volumes are measured at the same temperature and pressure in [Stoichiometry of Gaseous Substances, Mixtures, and Reactions](https://openstax.org/books/chemistry-2e/pages/9-3-stoichiometry-of-gaseous-substances-mixtures-and-reactions). Britannica also summarizes Gay-Lussac's finding that hydrogen and oxygen combine by volume in a ratio to form water in [Gay-Lussac's law of combining volumes](https://www.britannica.com/science/Gay-Lussacs-law-of-combining-volumes).

The important word is **gas**. Coefficients can be read directly as volume ratios only for substances in the gas phase and only when those volumes are compared at the same temperature and pressure.

## Read Water as Vapor

Look at the balanced equation for forming water vapor.

```math
2\mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow 2\mathrm{H_2O(g)}
```

The coefficients $$2:1:2$$ mean $$2$$ volume parts of hydrogen react with $$1$$ volume part of oxygen to form $$2$$ volume parts of water vapor.

Visible text: The coefficients mean volume parts of hydrogen react with volume part of oxygen to form volume parts of water vapor.

```math
V_{\mathrm{H_2}} : V_{\mathrm{O_2}} : V_{\mathrm{H_2O}} = 2:1:2
```

So, if every gas is measured at the same temperature and pressure, $$200\ \mathrm{mL}$$ of hydrogen reacts with $$100\ \mathrm{mL}$$ of oxygen to produce $$200\ \mathrm{mL}$$ of water vapor. If the water has condensed into liquid, the liquid volume is no longer read directly from this gas ratio.

Visible text: So, if every gas is measured at the same temperature and pressure, of hydrogen reacts with of oxygen to produce of water vapor. If the water has condensed into liquid, the liquid volume is no longer read directly from this gas ratio.

## Gas Volume Meter

Choose a reaction below. Each tube represents one volume part of gas at the same temperature and pressure.

Component: CombiningVolumesLab
Props:
- title: Gas Volume Meter
- description: Compare tube heights to see reaction coefficients as gas volume ratios.
- labels: {
chooseMode: "Choose gas reaction",
reactants: "Reactants",
products: "Products",
reactionView: "Law of combining volumes gas model",
ratioLabel: "Volume ratio",
exampleLabel: "Scaled example",
volumeUnit: "vol",
modes: {
"water-vapor": {
tab: "Steam",
tabLabel: "Water vapor formation",
helperCaption: (
<>
Two parts of $$\mathrm{H_2}$$ meet one part of{" "}
$$\mathrm{O_2}$$ and form two parts of{" "}
$$\mathrm{H_2O(g)}$$.
</>
),
ratio: (
<>
$$\mathrm{H_2:O_2:H_2O} = 2:1:2$$.
</>
),
example: (
<>
$$200\ \mathrm{mL}$$ of{" "}
$$\mathrm{H_2}$$ needs{" "}
$$100\ \mathrm{mL}$$ of{" "}
$$\mathrm{O_2}$$.
</>
),
},
"ammonia-synthesis": {
tab: "Ammonia",
tabLabel: "Ammonia formation",
helperCaption: (
<>
One part of $$\mathrm{N_2}$$ needs three parts of{" "}
$$\mathrm{H_2}$$ to form two parts of{" "}
$$\mathrm{NH_3}$$.
</>
),
ratio: (
<>
$$\mathrm{N_2:H_2:NH_3} = 1:3:2$$.
</>
),
example: (
<>
$$1\ \mathrm{L}$$ of{" "}
$$\mathrm{N_2}$$ needs{" "}
$$3\ \mathrm{L}$$ of{" "}
$$\mathrm{H_2}$$.
</>
),
},
"ammonia-decomposition": {
tab: "Decompose",
tabLabel: "Ammonia decomposition",
helperCaption: (
<>
Two parts of $$\mathrm{NH_3}$$ break into three
parts of $$\mathrm{H_2}$$ and one part of{" "} ... [truncated; 1471 chars]
  Visible text: {
chooseMode: "Choose gas reaction",
reactants: "Reactants",
products: "Products",
reactionView: "Law of combining volumes gas model",
ratioLabel: "Volume ratio",
exampleLabel: "Scaled example",
volumeUnit: "vol",
modes: {
"water-vapor": {
tab: "Steam",
tabLabel: "Water vapor formation",
helperCaption: (
<>
Two parts of meet one part of{" "}
 and form two parts of{" "}
.
</>
),
ratio: (
<>
.
</>
),
example: (
<>
 of{" "}
 needs{" "}
 of{" "}
.
</>
),
},
"ammonia-synthesis": {
tab: "Ammonia",
tabLabel: "Ammonia formation",
helperCaption: (
<>
One part of needs three parts of{" "}
 to form two parts of{" "}
.
</>
),
ratio: (
<>
.
</>
),
example: (
<>
 of{" "}
 needs{" "}
 of{" "}
.
</>
),
},
"ammonia-decomposition": {
tab: "Decompose",
tabLabel: "Ammonia decomposition",
helperCaption: (
<>
Two parts of break into three
parts of and one part of{" "} ... [truncated; 1471 chars]

The model is not comparing mass. It compares gas volume. Because the temperature and pressure are the same, each volume part follows a coefficient in the balanced reaction.

## Ammonia Decomposition Example

Suppose $$600\ \mathrm{mL}$$ of ammonia gas decomposes completely at the same temperature and pressure. The balanced equation is:

Visible text: Suppose of ammonia gas decomposes completely at the same temperature and pressure. The balanced equation is:

```math
2\mathrm{NH_3(g)} \rightarrow 3\mathrm{H_2(g)} + \mathrm{N_2(g)}
```

The coefficients give this volume ratio:

```math
V_{\mathrm{NH_3}} : V_{\mathrm{H_2}} : V_{\mathrm{N_2}} = 2:3:1
```

Because $$2$$ parts of ammonia equal $$600\ \mathrm{mL}$$, $$1$$ volume part equals $$300\ \mathrm{mL}$$.

Visible text: Because parts of ammonia equal , volume part equals .

```math
\begin{aligned}
V_{\mathrm{H_2}} &= 3 \times 300 = 900\ \mathrm{mL} \\
V_{\mathrm{N_2}} &= 1 \times 300 = 300\ \mathrm{mL}
\end{aligned}
```

So decomposing $$600\ \mathrm{mL}$$ of ammonia gas produces $$900\ \mathrm{mL}$$ of hydrogen gas and $$300\ \mathrm{mL}$$ of nitrogen gas.

Visible text: So decomposing of ammonia gas produces of hydrogen gas and of nitrogen gas.

## Where the Shortcut Stops

This law is useful, but its conditions must stay visible.

- The reaction equation must already be balanced.
- The substances being compared must be gases.
- All gas volumes must be measured at the same temperature and pressure.
- If a product turns into a liquid or solid, that product volume is not read directly from the gas coefficients.

With those conditions in place, reaction coefficients do more than count moles. For gases under the same conditions, they also give a quick map of relative volumes.