If 20 grams of sulfur reacts with an excess of oxygen, what volume of the product SO2 would be collected at 108.5 kPa and 15 degrees Celseius?
S + O2 ==> SO2
mol S = grams/molar mass = ?
Convert mols S to mols SO2 using the coeficients in the balanced equation. In this case mols S = mols SO2.
Then PV = nRT and solve for V at the conditions listed.
To calculate the volume of SO2 produced, we need to apply the ideal gas law equation:
PV = nRT
Where:
- P is the pressure of the gas (given as 108.5 kPa)
- V is the volume of the gas (what we want to calculate)
- n is the number of moles of the gas (what we need to find)
- R is the ideal gas constant (0.0821 L·atm/(K·mol))
- T is the temperature in Kelvin (which we need to convert)
First, let's determine the number of moles of sulfur dioxide (SO2) produced. To do that, we have to know the balanced chemical equation for the reaction and the molar mass of sulfur.
The balanced equation for the reaction between sulfur and oxygen to form sulfur dioxide (SO2) is:
S + O2 -> SO2
Since sulfur has a molar mass of approximately 32.07 g/mol, we can find the number of moles of sulfur:
moles of sulfur = mass of sulfur / molar mass of sulfur
Given that we have 20 grams of sulfur:
moles of sulfur = 20 g / 32.07 g/mol
Now we can calculate the number of moles of SO2. Since the reaction is 1:1, the number of moles of SO2 will be the same as the moles of sulfur.
Now, let's convert the temperature from Celsius to Kelvin:
T(Kelvin) = T(Celsius) + 273.15
Given that the temperature is 15 degrees Celsius:
T(Kelvin) = 15 + 273.15
Now we have all the values needed to calculate the volume of SO2 using the ideal gas law equation:
PV = nRT
V = (nRT) / P
Substituting in the values:
V = (moles of SO2 x R x T) / P
Plugging in the numbers, where:
- moles of SO2 is the same as the moles of sulfur calculated earlier
- R = 0.0821 L·atm/(K·mol)
- T is the temperature in Kelvin (calculated earlier)
- P = 108.5 kPa
Calculate the volume of SO2 using the given values.