Gas Stoichiometry: Videos & Practice Problems

Gas stoichiometry involves calculating the relationships between gases in balanced chemical equations. Using the ideal gas law, PV=nRT, we can determine moles of gases from given pressure, volume, and temperature. Transitioning from known to unknown quantities requires a mole-to-mole comparison based on coefficients in the balanced equation. This process is essential for converting moles of reactants to products, facilitating accurate predictions of yields in gas-evolution reactions.

Gas Stoichiometry

Gas stoichiometry focuses on the numerical relationships between reactants and products in chemical reactions that produce gases. It builds upon the principles of stoichiometry, which involves using balanced chemical equations to relate the quantities of different substances involved in a reaction.

To perform gas stoichiometry calculations, a stoichiometric chart is utilized. This chart allows you to take a known quantity of one compound, often a gas, and use it to calculate the unknown quantity of another compound in the balanced equation. The balanced chemical equation provides the necessary ratios of the reactants and products, which are essential for these calculations.

Understanding gas stoichiometry is crucial for accurately predicting the outcomes of reactions involving gases, as it helps in determining how much of each substance will be produced or consumed. This knowledge is particularly important in fields such as chemistry, environmental science, and engineering, where gas reactions are common.

Gas Stoichiometry

In stoichiometry, understanding the relationships between reactants and products is crucial for solving problems involving chemical reactions. For instance, consider the reaction where 4 moles of solid silver react with 1 mole of oxygen gas to produce 2 moles of silver oxide solid. To determine the grams of silver oxide produced, we can utilize the ideal gas law, represented by the equation:

\[PV = nRT\]

Here, \(P\) stands for pressure, \(V\) for volume, \(n\) for moles, \(R\) is the ideal gas constant, and \(T\) is temperature. By rearranging this equation, we can isolate the number of moles of the gas:

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

With the provided values of pressure, volume, and temperature, we can calculate the moles of oxygen gas available for the reaction. This step is essential as it allows us to transition from the known quantities (pressure, volume, temperature) to the moles of the reactant.

Once we have the moles of oxygen, we can make a stoichiometric leap to find the moles of silver oxide produced. This leap involves using the coefficients from the balanced chemical equation to perform a mole-to-mole conversion. For example, if we know the moles of oxygen, we can use the ratio from the balanced equation to find the moles of silver oxide:

\[\text{Moles of Silver Oxide} = \text{Moles of Oxygen} \times \frac{2 \text{ moles of Silver Oxide}}{1 \text{ mole of Oxygen}}\]

After determining the moles of silver oxide, converting this quantity into grams is straightforward using the molar mass of silver oxide. This process highlights the importance of stoichiometric relationships in chemical reactions, allowing us to predict the amounts of products formed from given reactants.

For those unfamiliar with stoichiometric charts, revisiting foundational concepts in stoichiometry can provide clarity and enhance problem-solving skills in gas stoichiometry scenarios.

Gas Stoichiometry Example 1

To determine the mass of silver oxide produced from the reaction of oxygen gas with excess solid silver, we start by using the ideal gas law, which relates pressure, volume, and temperature to the number of moles of a gas. The formula for calculating moles is given by:

n = \frac{PV}{RT}

In this scenario, we are provided with the volume of oxygen gas (384 mL), the pressure (736 mmHg), and the temperature (25°C). First, we need to convert the pressure from millimeters of mercury to atmospheres, knowing that 1 atmosphere equals 760 mmHg:

P = \frac{736 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.9684 \text{ atm}

Next, we convert the volume from milliliters to liters:

V = 0.384 \text{ L}

We also convert the temperature to Kelvin by adding 273.15:

T = 25 + 273.15 = 298.15 \text{ K}

Now, substituting these values into the ideal gas law equation allows us to calculate the moles of oxygen gas:

n = \frac{(0.9684 \text{ atm})(0.384 \text{ L})}{(0.0821 \text{ L atm/(mol K)})(298.15 \text{ K})} \approx 0.015199 \text{ moles of } O_2

With the moles of oxygen calculated, we can now perform a stoichiometric conversion to find the moles of silver oxide produced. The balanced chemical equation for the reaction is:

2 Ag + O_2 \rightarrow 2 Ag_2O

This indicates that 1 mole of oxygen gas produces 2 moles of silver oxide. Thus, we set up the mole ratio:

\text{Moles of } Ag_2O = 0.015199 \text{ moles } O_2 \times \frac{2 \text{ moles } Ag_2O}{1 \text{ mole } O_2} = 0.030398 \text{ moles } Ag_2O

Next, we convert moles of silver oxide to grams. The molar mass of silver oxide (Ag₂O) is calculated as follows:

\text{Molar mass of } Ag_2O = (2 \times 107.87 \text{ g/mol}) + (16.00 \text{ g/mol}) = 230.74 \text{ g/mol}

Now, we can find the mass of silver oxide produced:

\text{Mass of } Ag_2O = 0.030398 \text{ moles } Ag_2O \times 230.74 \text{ g/mol} \approx 7.0 \text{ g}

In this calculation, we maintain significant figures based on the least precise measurement, which in this case is the temperature (25°C, 2 significant figures). Therefore, the final answer is that 7.0 grams of silver oxide are produced from the reaction.

Understanding the theoretical yield is crucial in stoichiometry, as it allows for the comparison of calculated amounts to actual experimental results. For further insights into theoretical yield and its implications, additional resources on the topic are recommended.

The oxidation of phosphorus can be represented by the following equation:

                       P₄ (s)  +  5 O₂ (g) → 2 P₂O₅ (g)

If 1.85 L of diphosphorus pentoxide form at a temperature of 50.0 ºC and 1.12 atm, what is the mass (in g) of phosphorus that reacted?

• A. 4.84 g
• B. 7.02 g
• C. 10.9 g
• D. 12.4 g
• E. 17.7 g

Answer: (A)

Determine the mass (in grams) of water formed when 15.3 L NH₃ (at 298 K and 1.50 atm) is reacted with 21.7 L of O₂ (at 323 K and 1.1 atm).


4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)

• A. 12 g
• B. 19 g
• C. 25 g
• D. 38 g
• E. 51 g

Answer: (B)

Here’s what students ask on this topic:

What is gas stoichiometry and how is it used in chemical reactions?

Gas stoichiometry involves calculating the relationships between gases in balanced chemical equations. It uses the ideal gas law, $PV=nRT$, to determine the moles of gases from given pressure, volume, and temperature. By knowing the moles of a gas, we can use stoichiometric coefficients from the balanced equation to find the moles of other reactants or products. This process is essential for predicting yields in reactions that produce or consume gases.

How do you use the ideal gas law in gas stoichiometry calculations?

In gas stoichiometry, the ideal gas law $PV=nRT$ is used to find the number of moles of a gas. Given the pressure (P), volume (V), and temperature (T) of a gas, you can rearrange the equation to solve for moles (n): $n=\frac{PV}{RT}$. Once you have the moles, you can use stoichiometric coefficients from the balanced chemical equation to find the moles of other substances involved in the reaction.

What is the stoichiometric chart and how is it used in gas stoichiometry?

The stoichiometric chart is a tool used to visualize and organize the steps in stoichiometric calculations. It helps in converting given quantities (like grams or volume) to moles, and then using mole-to-mole ratios from the balanced equation to find unknown quantities. In gas stoichiometry, the chart incorporates the ideal gas law to convert between volume, pressure, temperature, and moles of gases. This systematic approach ensures accurate calculations and helps in understanding the relationships between reactants and products in a chemical reaction.

How do you convert between moles of a gas and grams of a solid in a chemical reaction?

To convert between moles of a gas and grams of a solid in a chemical reaction, follow these steps: 1) Use the ideal gas law $PV=nRT$ to find the moles of the gas. 2) Use the stoichiometric coefficients from the balanced chemical equation to convert moles of the gas to moles of the solid. 3) Finally, use the molar mass of the solid to convert moles to grams. This process ensures accurate conversion between different states of matter in a reaction.

What role do stoichiometric coefficients play in gas stoichiometry?

Stoichiometric coefficients in a balanced chemical equation represent the mole ratios of reactants and products. In gas stoichiometry, these coefficients are crucial for converting between moles of different substances. For example, if the equation shows that 1 mole of gas A reacts with 2 moles of gas B to produce 1 mole of gas C, the coefficients (1:2:1) guide the mole-to-mole conversions. This allows for accurate predictions of the amounts of reactants needed or products formed in a gas-evolution reaction.