Sulfur trioxide can be produced by the reaction in the gaseous phase, sulphur dioxide with molecular oxygen in the presence of a catalyst. If 128 g
branches of SO2 were placed in the presence of 16 grams of O2, that amount of SO3 will be theoretically formed?
To find the theoretical amount of SO3 formed, we need to determine the limiting reactant in the reaction between SO2 and O2.
Step 1: Write the balanced chemical equation:
2 SO2 + O2 -> 2 SO3
Step 2: Calculate the number of moles for each reactant:
Molar mass of SO2 = 32 g/mol
Number of moles of SO2 = mass of SO2 / molar mass of SO2 = 128 g / 32 g/mol = 4 mol SO2
Molar mass of O2 = 32 g/mol
Number of moles of O2 = mass of O2 / molar mass of O2 = 16 g / 32 g/mol = 0.5 mol O2
Step 3: Use the mole ratio from the balanced equation to determine the limiting reactant:
From the balanced equation, we can see that the ratio of SO2 to O2 is 2:1. This means that for every 2 moles of SO2, we need 1 mole of O2. However, in this case, we have 4 moles of SO2 and only 0.5 moles of O2. Since we don't have enough O2 to react with all the SO2, O2 is the limiting reactant.
Step 4: Calculate the theoretical amount of SO3 formed:
From the balanced equation, we know that for every 2 moles of SO2, we can produce 2 moles of SO3. Since O2 is the limiting reactant and reacts in a 1:1 ratio with SO3, we will produce an equal amount of SO3. Therefore, the number of moles of SO3 formed will be equal to the number of moles of O2 used, which is 0.5 mol.
Step 5: Calculate the mass of SO3 formed:
Molar mass of SO3 = 80 g/mol
Mass of SO3 = number of moles of SO3 × molar mass of SO3 = 0.5 mol × 80 g/mol = 40 g
Thus, theoretically, 40 grams of SO3 will be formed.