In this question, we have the following reaction:
2 SO2 + O2 -> 2 SO3, this reaction is already balanced
Considering molar ratio, we have:
2 SO2 = 1 O2
2 SO2 = 2 SO3
1 O2 = 2 SO3
Now, we need to find the limiting reactant, we will do it by checking the number of moles of both SO2 and O2 and compare it with the molar ratio
Let's start with SO2, molar ratio = 64.066g/mol
64.066g = 1 mol
35.0g = x moles
x = 0.546 moles of SO2 in 35 grams
According to the molar ratio, if we have 0.546 moles of SO2, we need half of it of O2, therefore:
0.546/2 = 0.273 moles of O2 is required
We need to check the number of moles of O2 to analyze if we have less or more moles than we actually need, O2 molar mass = 32g/mol
32g = 1 mol
35g = x moles
x = 1.094 moles of O2
If we have 1.094 moles and we need only 0.273 moles, this means that O2 is in excess and SO2 is the limiting reactant
Using the limiting reactant we can find the mass produced of SO3, according to the molar ratio, we will have the same number of moles from both SO2 and SO3, 0.546 moles
Now to find the mass, we use the number of moles, 0.546 and also the molar mass of SO3, 80.06g/mol
80.06g = 1 mol
x grams = 0.546 moles
x = 43.71 grams of SO3
Limiting reactant = SO2
Excess reactant = O2
SO3 formed = 43.71 grams