If 24.0 L of O2 burns completely a sample of carbon disulfide, how many moles of sulfur dioxide (SO2) will be produced at 15 degrees Celsius at a pressure of 1.3 atm?
CS2 + 3O2 -> CO2 + 2SO2
See the initial post that I commented on.
To find the number of moles of sulfur dioxide (SO2) produced when a given amount of oxygen (O2) is consumed, we need to use stoichiometry, which is the numerical relationship between the amounts of reactants and products in a chemical reaction.
First, we need to convert the volume of O2 gas (24.0 L) to moles using the ideal gas law equation:
PV = nRT
Where:
P = pressure (1.3 atm),
V = volume (24.0 L),
n = number of moles of gas,
R = ideal gas constant (0.0821 L·atm/(mol·K)),
T = temperature (15 degrees Celsius = 288.15 K)
Rearranging the ideal gas law equation, we can solve for n:
n = (PV) / (RT)
Substituting the known values:
n = (1.3 atm * 24.0 L) / (0.0821 L·atm/(mol·K) * 288.15 K)
Simplifying, we find:
n = 1.56 moles of O2
From the balanced chemical equation:
CS2 + 3O2 → CO2 + 2SO2
We can see that 3 moles of O2 react with 2 moles of SO2.
Using a ratio, we can calculate the number of moles of SO2 produced:
(2 moles SO2 / 3 moles O2) * 1.56 moles O2 = 1.04 moles SO2
Therefore, approximately 1.04 moles of sulfur dioxide (SO2) will be produced.