How many liters of sulphur trioxide are formed when 4800cm3 of sulphur dioxide is burned in air?
4.8L
4.8
2SO2 + O2 ==> 2SO3
4800 cc SO2 will produce 4800 cc SO3
4.8 litre
Exercise 4.9
4.8
4.8
How many liters of sulphur trioxide are formed when 4800 cm3 of sulphur dioxide in burned in air
To find out how many liters of sulphur trioxide are formed when 4800cm3 of sulphur dioxide is burned in air, we need to use stoichiometry and the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction is:
2 SO2 + O2 -> 2 SO3
From the balanced equation, we can see that 2 moles of sulphur dioxide (SO2) reacts to form 2 moles of sulphur trioxide (SO3).
To find the number of moles of SO2, we can use the ideal gas equation:
PV = nRT
Where:
P = pressure (assuming it's constant, it cancels out in the calculation)
V = volume of the gas in liters (4800 cm3 = 4800/1000 = 4.8 L)
n = number of moles of the gas (what we want to find)
R = ideal gas constant (constant value)
T = temperature (assuming it's constant, it cancels out in the calculation)
Rearranging the ideal gas equation, we can solve for n:
n = PV/RT
Since the pressure and temperature are assumed to be constant and cancel out in the calculation, we can simplify the equation to:
n = V/R
Now, we'll find the number of moles of SO2:
n(SO2) = 4.8 L / (R)
To find the number of moles of SO3 formed, we use the stoichiometry relationship from the balanced equation:
n(SO3) = n(SO2)
Now, we need to convert moles of SO3 to liters. Since 1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP), we can use this information to find the volume:
V(SO3) = n(SO3) * 22.4 L/mol
Finally, we can substitute the value of n(SO3) with n(SO2):
V(SO3) = n(SO2) * 22.4 L/mol
To calculate the answer, we need the value of R (ideal gas constant) and the molar mass of SO2. Once we have those values, we can go ahead and calculate the result.