Sulfur dioxide (SO2) reacts with oxygen gas to make sulfur trioxide (SO3).
How many grams of oxygen are used to produce 8.11 grams of sulfur trioxide?
2SO2 + O2 ==> 2SO3
mols SO3 needed = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols SO3 to mols O2.
Now convert mols O2 to grams. g O2 = mols O2 x molar mass O2.
To solve this problem, you need to use the concept of stoichiometry, which is the study of the quantitative relationships between reactants and products in a chemical reaction.
The balanced chemical equation for the reaction between sulfur dioxide and oxygen gas is:
2 SO2 + O2 → 2 SO3
From this equation, we can see that for every 1 mole of sulfur trioxide (SO3) produced, we need 1 mole of oxygen gas (O2).
To determine the amount of oxygen gas required to produce 8.11 grams of sulfur trioxide, we need to perform the following steps:
Step 1: Convert the given mass of SO3 to moles.
To do this, we need to divide the mass of SO3 by its molar mass. According to the periodic table, the molar mass of sulfur trioxide (SO3) is 80.06 grams/mole.
moles of SO3 = mass of SO3 / molar mass of SO3
moles of SO3 = 8.11 g / 80.06 g/mol
moles of SO3 = 0.1013 mol
Step 2: Use the stoichiometric ratio from the balanced equation to determine the moles of oxygen gas required.
From the balanced equation, we know that the stoichiometric ratio between SO3 and O2 is 1:1. Therefore, the moles of oxygen gas (O2) required will be equal to the moles of sulfur trioxide (SO3).
moles of O2 = moles of SO3
moles of O2 = 0.1013 mol
Step 3: Convert the moles of oxygen gas to grams.
To do this, we need to multiply the moles of O2 by its molar mass. According to the periodic table, the molar mass of oxygen gas (O2) is 32.00 grams/mole.
mass of O2 = moles of O2 × molar mass of O2
mass of O2 = 0.1013 mol × 32.00 g/mol
mass of O2 = 3.243 g
Therefore, 3.243 grams of oxygen gas are required to produce 8.11 grams of sulfur trioxide.