For the reaction represented by the equation SO3+H2O–>H22SO4 how many grams of sulfuric acid (H22SO4) can be produced from 197 g of water and 460g of sulfer trioxide?
The correct answer and explanation is :
To determine how many grams of sulfuric acid (H₂SO₄) can be produced from the given amounts of water (H₂O) and sulfur trioxide (SO₃), we need to follow the steps of stoichiometry. Let’s break it down step-by-step.
1. Write the balanced chemical equation:
The balanced equation for the reaction is:
[
SO_3 + H_2O \rightarrow H_2SO_4
]
This indicates that one mole of sulfur trioxide (SO₃) reacts with one mole of water (H₂O) to produce one mole of sulfuric acid (H₂SO₄).
2. Calculate the molar masses of the substances involved:
- Molar mass of H₂O:
H₂O consists of 2 hydrogen atoms and 1 oxygen atom.
Molar mass of H₂O = (2 × 1) + (1 × 16) = 18 g/mol. - Molar mass of SO₃:
SO₃ consists of 1 sulfur atom and 3 oxygen atoms.
Molar mass of SO₃ = (1 × 32) + (3 × 16) = 80 g/mol. - Molar mass of H₂SO₄:
H₂SO₄ consists of 2 hydrogen atoms, 1 sulfur atom, and 4 oxygen atoms.
Molar mass of H₂SO₄ = (2 × 1) + (1 × 32) + (4 × 16) = 98 g/mol.
3. Determine the limiting reactant:
We are given 197 g of water (H₂O) and 460 g of sulfur trioxide (SO₃). Let’s calculate how many moles of each reactant we have:
- Moles of H₂O:
[
\text{Moles of H₂O} = \frac{\text{Mass of H₂O}}{\text{Molar mass of H₂O}} = \frac{197 \, \text{g}}{18 \, \text{g/mol}} = 10.94 \, \text{mol}
] - Moles of SO₃:
[
\text{Moles of SO₃} = \frac{\text{Mass of SO₃}}{\text{Molar mass of SO₃}} = \frac{460 \, \text{g}}{80 \, \text{g/mol}} = 5.75 \, \text{mol}
]
From the balanced equation, we see that the stoichiometric ratio between SO₃ and H₂O is 1:1, so we need an equal number of moles of both reactants. Since we have fewer moles of SO₃ (5.75 mol) than H₂O (10.94 mol), SO₃ is the limiting reactant.
4. Calculate the moles of H₂SO₄ produced:
Since SO₃ is the limiting reactant and the stoichiometric ratio is 1:1 (1 mole of SO₃ produces 1 mole of H₂SO₄), we will produce 5.75 moles of H₂SO₄.
5. Convert moles of H₂SO₄ to grams:
Now, to find the mass of sulfuric acid produced, we multiply the moles of H₂SO₄ by its molar mass:
[
\text{Mass of H₂SO₄} = \text{Moles of H₂SO₄} \times \text{Molar mass of H₂SO₄}
]
[
\text{Mass of H₂SO₄} = 5.75 \, \text{mol} \times 98 \, \text{g/mol} = 563.5 \, \text{g}
]
Conclusion:
The maximum amount of sulfuric acid (H₂SO₄) that can be produced from 197 g of water and 460 g of sulfur trioxide is 563.5 grams.
Explanation:
- We first determined the molar amounts of water and sulfur trioxide.
- We identified sulfur trioxide (SO₃) as the limiting reactant since there were fewer moles of SO₃ compared to water.
- Using the limiting reactant, we calculated the moles of H₂SO₄ that can be produced and converted that into grams.
This gives us the final mass of sulfuric acid, which is 563.5 grams.