To calculate the total volume of gas in the container at the end of the reaction, we first need to determine the balanced chemical equation for the reaction between SO2 and oxygen.
The balanced chemical equation is:
2 SO2 + O2 -> 2 SO3
From the balanced equation, we can see that 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3.
Given that 96 g of SO2 is provided, we need to convert this to moles to determine how many moles of SO2 we have:
Molar mass of SO2 = 32.06 g/mol
Number of moles of SO2 = 96 g / (32.06 g/mol) = 2.9957 mol (rounded to 4 decimal places)
Since the reaction ratio is 2 moles of SO2 to 1 mole of O2, we can say that the number of moles of O2 consumed is half the number of moles of SO2:
Number of moles of O2 consumed = 2.9957 mol / 2 = 1.4978 mol (rounded to 4 decimal places)
Now, we need to determine the total number of moles of gas at the end of the reaction. Since the volume of gas is directly proportional to the number of moles, we can use the ideal gas law equation to calculate the volume:
PV = nRT
Where:
P is the pressure (which is assumed to be at STP, so 1 atm)
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature in Kelvin (273 K at STP)
Rearranging the equation to solve for V:
V = nRT / P
Substituting the values, we get:
V = (1.4978 mol)(0.0821 L·atm/mol·K)(273 K) / 1 atm = 33.11 L (rounded to 2 decimal places)
Therefore, the total volume of gas in the container at the end of the reaction is 33.11 L.