Consider the following chemical equation.
SO2(g) + O2(g) → SO3(g)
What volume of sulfur dioxide gas reacts with 37.5 L of O2? The reaction conditions are 875°C and 1.00 atm pressure.
To find the volume of sulfur dioxide gas that reacts with 37.5 L of oxygen gas (O2), we need to use stoichiometry.
1. Start by balancing the chemical equation:
SO2(g) + O2(g) → SO3(g)
The balanced equation already provides the stoichiometric ratio between SO2 and O2, which is 1:1.
2. Convert the given volume of oxygen gas (O2) to the number of moles:
Using the ideal gas law equation, PV = nRT, we can convert the volume to moles.
n(O2) = (PV) / (RT)
n(O2) = (1.00 atm * 37.5 L) / (0.0821 L·atm/K·mol * 875°C + 273.15)
3. Use the stoichiometry from the balanced equation to determine the number of moles of SO2 reacting:
Since the stoichiometric ratio between SO2 and O2 is 1:1, the number of moles of SO2 will also be equal to n(O2) calculated in step 2.
4. Finally, convert the number of moles of SO2 to volume:
Using the ideal gas law equation, PV = nRT, we can convert the moles of SO2 to volume.
V(SO2) = (n(SO2) * RT) / P
V(SO2) = (n(O2) * RT) / P
Note: In step 2, make sure to convert the temperature into Kelvin by adding 273.15.
Plug in the values into the equation to calculate the volume of sulfur dioxide gas (SO2) that reacts with 37.5 L of oxygen gas.