2so2+o2-2so3 if 96g so2 is added to 2 moles of oxygen at STP calculate the volume of so3 that is formed
To calculate the volume of SO3 formed, we need to use the stoichiometry of the reaction and the ideal gas law.
First, let's write the balanced chemical equation for the reaction:
2 SO2 + O2 → 2 SO3
From the balanced equation, we can see that for every 2 moles of SO2 reacted, 2 moles of SO3 are formed. Therefore, the molar ratio is 2:2 or 1:1.
Given that 96 g of SO2 is added, we need to convert it to moles using its molar mass. The molar mass of SO2 is:
32.06 g/mol (sulfur) + 2 * 16.00 g/mol (oxygen) = 64.06 g/mol
Moles of SO2 = Mass of SO2 / Molar mass of SO2
= 96 g / 64.06 g/mol
≈ 1.50 mol
Since the molar ratio of SO2 to SO3 is 1:1, the number of moles of SO3 formed will also be 1.50 mol.
Now we can use the ideal gas law to calculate the volume of SO3 at STP (Standard Temperature and Pressure). STP is defined as 1 atm pressure and 273.15 K temperature.
The ideal gas law equation is:
PV = nRT
where:
P = Pressure (1 atm)
V = Volume
n = number of moles (1.50 mol)
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature in Kelvin (273.15 K)
Solving for V, we get:
V = nRT / P
= (1.50 mol) * (0.0821 L·atm/mol·K) * (273.15 K) / (1 atm)
≈ 33.6 L
Therefore, approximately 33.6 liters of SO3 will be formed when 96 grams of SO2 react with 2 moles of oxygen at STP.