An excces of sulfur dioxide was reacted with 0.150 g of oxigen gas. A mass of 0.725 g of sulfur trioxide was recovered
96.5%
To find the stoichiometry of the reaction and determine the molar ratio between sulfur dioxide (SO2) and sulfur trioxide (SO3), we need to use the given masses of reactants and products.
First, let's convert the given masses of reactants and products to moles. To do this, we divide the given mass by the molar mass of each substance.
The molar mass of sulfur dioxide (SO2) is calculated as follows:
S: 32.07 g/mol
O: 16.00 g/mol (two oxygen atoms)
Total molar mass of SO2: 32.07 + 2(16.00) = 64.07 g/mol
The moles of sulfur dioxide (SO2) can be calculated as:
moles of SO2 = mass of SO2 / molar mass of SO2
moles of SO2 = mass of SO2 / 64.07 g/mol
Using the given mass of sulfur dioxide:
moles of SO2 = 0.150 g / 64.07 g/mol
Similarly, we can calculate the moles of oxygen gas (O2) using its molar mass:
O: 16.00 g/mol (one oxygen molecule)
Total molar mass of O2 = 2(16.00) = 32.00 g/mol
moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 0.150 g / 32.00 g/mol
Next, let's convert the given mass of sulfur trioxide (SO3) to moles. Repeat the same process with its molar mass:
S: 32.07 g/mol
O: 16.00 g/mol (three oxygen atoms)
Total molar mass of SO3: 32.07 + 3(16.00) = 80.07 g/mol
moles of SO3 = mass of SO3 / molar mass of SO3
moles of SO3 = 0.725 g / 80.07 g/mol
Now, we have the molar ratios between the reactants and the product. The balanced equation for the reaction is:
2 SO2 + O2 → 2 SO3
From the equation, we can see that the stoichiometric ratio between SO2 and SO3 is 2:2, which simplifies to a 1:1 ratio.
Finally, based on the given masses and the stoichiometric ratio, we can determine the excess reactant by comparing the mole ratios. If all the SO2 reacted exactly as per the stoichiometry, we should have the same number of moles of SO3. If there is a difference, it indicates which reactant is in excess.
Let's calculate the number of moles of SO2 and SO3:
moles of SO2 = 0.150 g / 64.07 g/mol
moles of SO3 = 0.725 g / 80.07 g/mol
As we can see, the moles of SO2 are less than the moles of SO3. Therefore, SO2 is the limiting reactant, and O2 is in excess.
I hope this explanation helps you understand how to determine the limiting reactant and excess reactant in a chemical reaction.