according to the balanced chemical equation 2SO2<--- 2 is a subscript...so anyway it's 2SO2+ O2 ----> 2SO3
How many liters of oxygen would be needed to produce 0.5 moles of sulfur trioxide, SO3?
At what pressure and temperature? STP?
0.5 mol SO3 will need 1/2 that or 0.25 mol O2.
1 mol O2 occupies 22.4L @ STP so 0.25 mol will occupy ......
yes its assuming at STP
To calculate the number of liters of oxygen needed to produce 0.5 moles of sulfur trioxide (SO3), we need to use the balanced chemical equation and the molar ratio between oxygen and sulfur trioxide.
The balanced chemical equation is:
2SO2 + O2 -> 2SO3
From this equation, we can see that for every 1 mole of oxygen (O2), 2 moles of sulfur trioxide (SO3) are produced. Therefore, the molar ratio between oxygen and sulfur trioxide is 1:2.
Given that you have 0.5 moles of sulfur trioxide, you can use this molar ratio to determine the moles of oxygen needed. Since the ratio is 1:2, you would need double the number of moles of sulfur trioxide for the moles of oxygen required.
Moles of oxygen = 2 * moles of sulfur trioxide = 2 * 0.5 = 1 mole of oxygen.
Now, to convert moles of oxygen to liters, we need to use the ideal gas law equation:
PV = nRT
Rearranging the equation:
V = (nRT) / P
Where:
V = volume in liters
n = number of moles of gas (oxygen in this case)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (assume room temperature, 298K)
P = pressure (assume atmospheric pressure, 1 atm)
Substituting the values into the equation, we have:
V = (1 mol * 0.0821 L·atm/(mol·K) * 298 K) / 1 atm
V = 24.44 liters
Therefore, to produce 0.5 moles of sulfur trioxide, you would need approximately 24.44 liters of oxygen.