how many liters of SO2 measured at STP are produced in the reaction of 6.27 kilogrrams with 234 Liters of O2 at 220 degrees celsius and 2 atm?
This is a limiting reagent (LR) problem and you know that because amounts are given for both reactants.
6.27 kg what?
PbS
2PbS + 3O2 ==> 2PbO + 2SO2
This is a limiting reagent (LR) problem and I work these the long way.
mols Pb S = 6270 g/molar mass = ?
mols O2. Use PV = nRT and solve for mols = n = ?
Using the coefficients in the balanced equation, convert mols PbS to mols SO2
Do the same and convert mols O2 to mols SO2.
It is likely that the two values will not agree which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reatent producing that value is the LR.
Using the smaller value, convert mols SO2 to L L = mols SO2 x (22.4 L/mol) = ?
To determine the number of liters of SO2 produced in the reaction, we need to use the given information and convert the given quantities to the appropriate units.
1. Start by converting the mass of SO2 from kilograms to grams:
6.27 kilograms = 6.27 × 1000 grams = 6270 grams
2. Next, calculate the number of moles of SO2. To do this, we need to use the molar mass of SO2, which is 64.06 g/mol:
Moles of SO2 = Mass of SO2 / Molar mass of SO2
Moles of SO2 = 6270 grams / 64.06 g/mol ≈ 97.80 moles
3. Now, we can use the balanced equation to determine the stoichiometric ratio of SO2 to O2:
2 SO2 + O2 → 2 SO3
Based on the equation, for every 2 moles of SO2, we need 1 mole of O2. Therefore, we can determine the number of moles of O2 used in the reaction. Since we have 234 liters of O2, we need to convert it to moles. To do this, we'll use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin
We need to convert the given temperature from Celsius to Kelvin:
Temperature in Kelvin = 220 degrees Celsius + 273.15 = 493.15 K
Now, let's calculate the number of moles of O2:
n = (PV) / (RT)
n = (2 atm × 234 L) / (0.0821 L.atm/mol.K × 493.15 K) ≈ 9.38 moles
4. Now, we can determine the number of moles of SO2 consumed in the reaction. Since the stoichiometric ratio is 2:1 for SO2 to O2, we divide the number of moles of O2 by 2:
Moles of SO2 consumed = Moles of O2 / 2
Moles of SO2 consumed ≈ 9.38 moles / 2 ≈ 4.69 moles
5. Finally, we can calculate the volume of SO2 produced using the ideal gas law equation. Since the reaction is measured at STP (standard temperature and pressure), we'll assume the temperature is 273.15 K and the pressure is 1 atm:
V = (nRT) / P
V = (4.69 moles × 0.0821 L.atm/mol.K × 273.15 K) / 1 atm ≈ 105.1 liters
Therefore, approximately 105.1 liters of SO2 measured at STP are produced in the reaction.