Gas Stoichiometry: Videos & Practice Problems

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.

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.

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.