SO2(g) + O2(g) → SO3(g)
What volume of sulfur dioxide gas reacts with 15.0 L of O2? The reaction conditions are 875°C and 1.00 atm pressure
Balance the equation.
2SO2 + O2 ==> 2SO3
A short cut to stoichiometry problems occurs when all are gases and at the same P and T. The shortcut is that you may use volume and mols interchangeably.
15.0 L O2 x (2 mol SO2/1 mol O2) = ? L SO2.
To determine the volume of sulfur dioxide gas that reacts with 15.0 L of O2, you need to use the balanced chemical equation and the principles of stoichiometry.
The balanced chemical equation for the reaction is:
2 SO2(g) + O2(g) → 2 SO3(g)
From the equation, we can see that 2 moles of sulfur dioxide react with 1 mole of oxygen gas to produce 2 moles of sulfur trioxide.
So, the stoichiometric ratio is 2 moles of SO2 : 1 mole of O2.
First, let's convert the given volume of O2 (15.0 L) to moles, using the ideal gas law equation:
P × V = n × R × T
Given:
Pressure (P) = 1.00 atm
Volume (V) = 15.0 L
Temperature (T) = 875°C = 875 + 273 = 1148 K (converted to Kelvin)
Gas constant (R) = 0.0821 L·atm/(mol·K) (ideal gas constant)
Plugging in the values, we can solve for the number of moles (n) of O2:
(1.00 atm) × (15.0 L) = n × (0.0821 L·atm/(mol·K)) × (1148 K)
n = (1.00 atm × 15.0 L) / (0.0821 L·atm/(mol·K) × 1148 K) = 0.1692 mol of O2
Now, using the stoichiometry of the balanced equation, we can find the amount of sulfur dioxide that reacts with this amount of oxygen.
From the stoichiometric ratio, 2 moles of SO2 react with 1 mole of O2.
So, we can write the proportion: 2 moles of SO2 / 1 mole of O2 = x moles of SO2 / 0.1692 mol of O2
Solving for x, we have:
x = (2 moles SO2 / 1 mole O2) × (0.1692 mol O2) = 0.3384 mol SO2
Finally, to find the volume of SO2, we can use the ideal gas law equation:
P × V = n × R × T
Given:
Pressure (P) = 1.00 atm
Temperature (T) = 875°C = 875 + 273 = 1148 K (converted to Kelvin)
Gas constant (R) = 0.0821 L·atm/(mol·K) (ideal gas constant)
Number of moles (n) = 0.3384 mol (from above calculation)
Plugging in the values, we can solve for the volume (V) of SO2:
(1.00 atm) × V = (0.3384 mol) × (0.0821 L·atm/(mol·K)) × (1148 K)
V = (0.3384 mol × 0.0821 L·atm/(mol·K) × 1148 K) / (1.00 atm) = 31.5 L
Therefore, the volume of sulfur dioxide gas that reacts with 15.0 L of O2 under the given conditions is 31.5 L.