If 96g SO2 is added to 2 moles of oxygen at STP,calculate the volume of SO3 that is formed
This is a limiting reagent problem.
2SO2 + O2 ==> 2SO3
How many mols SO2 do you have in 96 g. That's
mols = grams/molar mass = 96/64 = 1.5
1 mol SO2 produces 1 mol SO3 so you will get 1.5 mols SO3 IF there is enough oxygen.
From the equation 1 mol O2 will produce 2 mol SO3 so 2 mol O2 will produce 4 mols SO3. The LR is SO2 so you will get 1.5 mols SO3. 1 mol will occupy 22.4 L@ STP. So 1.5 mol SO3 @ STP will occupy .........L.
Getting answers
Calculate the mass and volume 2SO²+O²=2SO³
To calculate the volume of SO3 formed, we need to use the balanced chemical equation for the reaction between SO2 and O2 to form SO3:
2 SO2 + O2 → 2 SO3
Since we know the amount of SO2 (96g) and the starting amount of O2 (2 moles), we can use stoichiometry to determine the amount of SO3 formed.
1. Convert the mass of SO2 to moles:
First, find the molar mass of SO2 by adding up the atomic masses of its constituents:
S = 32.06 g/mol
O = 16.00 g/mol (2 oxygen atoms)
Molar mass of SO2 = 32.06 g/mol + 2 × 16.00 g/mol = 64.06 g/mol
Now, calculate the number of moles of SO2 by dividing the mass of SO2 by its molar mass:
moles of SO2 = 96g / 64.06 g/mol = 1.5 moles
2. Determine the limiting reactant:
Since the balanced equation shows that we need 2 moles of SO2 to react with 1 mole of O2, we can see that we have an excess of O2. This means O2 is the limiting reactant, and SO2 is in excess.
3. Calculate the moles of SO3 formed:
From the balanced equation, we know that for every 2 moles of SO2, 2 moles of SO3 are formed. Since we have 1.5 moles of SO2, we will have the same number of moles of SO3 formed:
moles of SO3 formed = 1.5 moles
4. Calculate the volume of SO3 at STP:
At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters. Therefore, the volume of SO3 formed can be calculated by multiplying the number of moles of SO3 by the molar volume at STP:
volume of SO3 formed = moles of SO3 formed × 22.4 liters/mole
Substituting the value we obtained earlier for the moles of SO3 formed:
volume of SO3 formed = 1.5 moles × 22.4 liters/mole
Calculating:
volume of SO3 formed = 33.6 liters
Therefore, the volume of SO3 formed when 96g of SO2 is added to 2 moles of oxygen at STP is 33.6 liters.