For the reaction represented by the equation SO3 + H2O -> H2SO4, how many grams of sulfuric acid can be produced from 320 g of water and 1064 g of sulfuric trioxide

For the reaction represented by the equation SO3 + H2O -> H2SO4, how many grams of sulfuric acid can be produced from 320 g of water and 1064 g of sulfuric trioxide?

The Correct Answer and Explanation is:

To solve this problem, we need to determine the limiting reagent in the reaction and then calculate how much sulfuric acid (H₂SO₄) can be produced from the given amounts of reactants. The balanced chemical equation is:

SO3+H2O→H2SO4SO_3 + H_2O \rightarrow H_2SO_4SO3​+H2​O→H2​SO4​

Step 1: Molar Masses

First, we need to know the molar masses of the compounds involved:

• Molar mass of H₂O (water) = 18.02 g/mol
• Molar mass of SO₃ (sulfur trioxide) = 80.06 g/mol
• Molar mass of H₂SO₄ (sulfuric acid) = 98.08 g/mol

Step 2: Moles of Reactants

Next, we calculate how many moles of each reactant are available.

Moles of H₂O:

Given that we have 320 g of water:moles of H2O=320 g18.02 g/mol≈17.75 mol\text{moles of H}_2O = \frac{320 \, \text{g}}{18.02 \, \text{g/mol}} \approx 17.75 \, \text{mol}moles of H2​O=18.02g/mol320g​≈17.75mol

Moles of SO₃:

Given that we have 1064 g of sulfur trioxide:moles of SO3=1064 g80.06 g/mol≈13.30 mol\text{moles of SO}_3 = \frac{1064 \, \text{g}}{80.06 \, \text{g/mol}} \approx 13.30 \, \text{mol}moles of SO3​=80.06g/mol1064g​≈13.30mol

Step 3: Identify the Limiting Reagent

From the balanced equation, we see that the mole ratio of SO₃ to H₂O is 1:1. This means that for every 1 mole of SO₃, 1 mole of H₂O is required to produce H₂SO₄.

• We have 17.75 moles of H₂O and 13.30 moles of SO₃.
• Since SO₃ is the limiting reagent (because we have fewer moles of SO₃), the reaction will be limited by the amount of SO₃.

Step 4: Calculate the Grams of H₂SO₄ Produced

According to the reaction, 1 mole of SO₃ produces 1 mole of H₂SO₄. Therefore, the moles of H₂SO₄ produced will be equal to the moles of SO₃.moles of H2SO4=13.30 mol\text{moles of H}_2SO_4 = 13.30 \, \text{mol}moles of H2​SO4​=13.30mol

Now, calculate the mass of sulfuric acid produced:mass of H2SO4=moles of H2SO4×molar mass of H2SO4\text{mass of H}_2SO_4 = \text{moles of H}_2SO_4 \times \text{molar mass of H}_2SO_4mass of H2​SO4​=moles of H2​SO4​×molar mass of H2​SO4​mass of H2SO4=13.30 mol×98.08 g/mol≈1303.26 g\text{mass of H}_2SO_4 = 13.30 \, \text{mol} \times 98.08 \, \text{g/mol} \approx 1303.26 \, \text{g}mass of H2​SO4​=13.30mol×98.08g/mol≈1303.26g

Final Answer:

The mass of sulfuric acid that can be produced is approximately 1303.26 g.