For the reaction SO3 plus H2O>>H2SO4, how many grams of sulfuric acid can be produced from 200.g of sulfuric trioxide and 100.g of water??
To determine the grams of sulfuric acid produced, we first need to identify the limiting reagent, which is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Let's start by calculating the number of moles of each reactant.
1) Sulfur trioxide (SO3):
Given mass of SO3 = 200 g
Molar mass of SO3 = 80.066 g/mol
Moles of SO3 = (given mass of SO3) / (molar mass of SO3)
= 200 g / 80.066 g/mol
≈ 2.4989 mol
2) Water (H2O):
Given mass of H2O = 100 g
Molar mass of H2O = 18.01528 g/mol
Moles of H2O = (given mass of H2O) / (molar mass of H2O)
= 100 g / 18.01528 g/mol
≈ 5.5488 mol
Next, we need to determine the mole ratio between SO3 and H2SO4 from the balanced chemical equation:
SO3 + H2O >> H2SO4
From the equation, we can see that 1 mole of SO3 produces 1 mole of H2SO4.
Now, let's compare the moles of SO3 and H2O to identify the limiting reagent.
The moles of SO3 is 2.4989 mol and the moles of H2O are 5.5488 mol.
From the comparison, we can see that there are more moles of H2O than SO3. Therefore, SO3 is the limiting reagent.
To calculate the moles of sulfuric acid produced from the limiting reagent (SO3), we use the mole ratio between SO3 and H2SO4, which is 1:1.
Moles of H2SO4 = Moles of SO3 = 2.4989 mol
Finally, to calculate the mass of sulfuric acid produced, we need to multiply the moles of H2SO4 by its molar mass:
Molar mass of H2SO4 = 98.07848 g/mol
Mass of H2SO4 = (moles of H2SO4) x (molar mass of H2SO4)
= 2.4989 mol × 98.07848 g/mol
≈ 245.06 g
Therefore, approximately 245.06 grams of sulfuric acid can be produced from 200 grams of sulfuric trioxide and 100 grams of water.