Sulfur trioxide dissolves in water, producing H2SO4. How much sulfuric acid can be produced from 12.1 mL of water (d=1.00g/mL) and 22.1 g of SO3?
To find the amount of sulfuric acid that can be produced from the given amounts of water and sulfur trioxide, we need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed first and limits the amount of product that can be formed.
First, let's calculate the number of moles of water and sulfur trioxide:
Moles of water = volume (in liters) × density × molar mass
= (12.1 mL ÷ 1000 mL/L) × (1.00 g/mL) ÷ (18.015 g/mol)
Moles of SO3 = mass ÷ molar mass
= 22.1 g ÷ 80.0642 g/mol
Now, we need to calculate the stoichiometry of the reaction. From the balanced chemical equation, we know that 1 mole of SO3 reacts with 1 mole of water to produce 1 mole of sulfuric acid (H2SO4).
Since the stoichiometric ratio is 1:1, the number of moles of sulfuric acid produced will be equal to the smaller of the moles of water and moles of SO3.
Next, we need to convert the moles of sulfuric acid to grams:
Mass of sulfuric acid = moles × molar mass
= (moles of SO3 or H2O) × (98.09 g/mol)
Finally, we can substitute the values into the equation to find the answer:
Mass of sulfuric acid = (minimum of moles of SO3 and H2O) × (98.09 g/mol)
Keep in mind that the given answer will be the smaller of the two amounts because the limiting reagent will determine the maximum amount of product formed.
To determine how much sulfuric acid can be produced, we need to find the limiting reagent in the reaction. This is the reactant that will be completely consumed and will determine the maximum amount of product that can be formed.
Let's start by finding the number of moles of water:
Number of moles of water = volume (in L) / molar volume
Molar volume of water = 18.015 g/mol (molar mass of water)
First, convert the volume of water from mL to L:
12.1 mL = 12.1/1000 = 0.0121 L
Number of moles of water = 0.0121 L / 18.015 g/mol = 0.000672 mol
Now, let's find the number of moles of sulfur trioxide (SO3):
Number of moles of SO3 = mass / molar mass
Molar mass of SO3 = 32.06 g/mol (molar mass of sulfur trioxide)
Number of moles of SO3 = 22.1 g / 32.06 g/mol = 0.689 mol
Next, let's determine the stoichiometry of the reaction. From the balanced equation, we know that 1 mole of SO3 reacts with 1 mole of water to produce 1 mole of sulfuric acid (H2SO4).
Since the mole ratio is 1:1, the limiting reagent is the reactant that has the smaller number of moles. In this case, water has the smaller number of moles (0.000672 mol), so it is the limiting reagent.
Now, let's calculate the amount of sulfuric acid that can be produced from the limiting reagent:
Number of moles of H2SO4 = number of moles of limiting reagent (water)
Number of moles of H2SO4 = 0.000672 mol
Finally, let's convert the number of moles of sulfuric acid to grams:
Mass of H2SO4 = number of moles of H2SO4 x molar mass of H2SO4
Molar mass of H2SO4 = 98.09 g/mol (molar mass of sulfuric acid)
Mass of H2SO4 = 0.000672 mol x 98.09 g/mol = 0.066 g
Therefore, 0.066 grams of sulfuric acid can be produced from 12.1 mL of water and 22.1 g of sulfur trioxide.