In the preparation of sulfuric acid, sulfur dioxide must be catalytically oxidized to sulfur trioxide. How much oxygen must be available to convert 50.0 L of sulfur dioxide according to this process?

To determine the amount of oxygen required for the conversion of 50.0 L of sulfur dioxide (SO2) to sulfur trioxide (SO3) through catalytic oxidation, we need to understand the stoichiometry of the reaction. The balanced chemical equation for this process is as follows:

2 SO2 + O2 -> 2 SO3

From the equation, we can see that two molecules of SO2 react with one molecule of O2 to form two molecules of SO3.

To find the amount of oxygen required, we can use the concept of mole-to-mole ratio. The ratio from the balanced equation tells us that two moles of SO2 react with one mole of O2. Therefore, we need to determine the number of moles of SO2 first.

To convert from liters (L) to moles (mol), we need to use the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)

Since the question does not provide pressure and temperature values, we'll assume standard conditions of 1 atm and 273 K.

Now let's calculate the number of moles of SO2:

n(SO2) = PV / RT

n(SO2) = (1 atm) * (50.0 L) / (0.0821 L·atm/mol·K * 273 K)
≈ 1.89 moles of SO2

According to the stoichiometry of the balanced equation, two moles of SO2 react with one mole of O2. Therefore, to determine the amount of O2 required, we multiply the number of moles of SO2 by the ratio of moles of O2 to moles of SO2:

n(O2) = (1.89 mol SO2) / 2
≈ 0.95 moles of O2

So, approximately 0.95 moles of oxygen (O2) are required to convert 50.0 L of sulfur dioxide (SO2) to sulfur trioxide (SO3) through catalytic oxidation.

Write the equation and balance it.

There is a shortcut you can use, since reactants and products are gases, that allows you to equation the coefficients to liters directly.