Question

For the reaction So3 + H2O -> H2SO4, how many grams of sulfur trioxide are required to produce 4.00 mol of sulfuric acid

Answer 1

Answer:

320 grams of sulfur trioxide are required to produce 4.00 mol of sulfuric acid.

Explanation:

The balanced reaction is:

SO₃ + H₂O → H₂SO₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

• SO₃: 1 mole
• H₂O: 1 mole
• H₂SO₄: 1 mole

Being the molar mass of each compound:

• SO₃: 80 g/mole
• H₂O: 18 g/mole
• H₂SO₄: 98 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

• SO₃: 1 mole* 80 g/mole= 80 grams
• H₂O: 1 mole* 18 g/mole= 18 grams
• H₂SO₄: 1 mole* 98 g/mole= 98 grams

Then you can apply the following rule of three: if 1 mole of sulfuric acid is produced by the reaction of 80 grams of sulfur trioxide, 4 moles of sulfuric acid is produced from how much mass of sulfur trioxide?

$mass of sulfur trioxide= \frac{4 moles of sulfuric acid* 80 grams of sulfur trioxide}{1 mole of sulfuric acid }$

mass of sulfur trioxide= 320 grams

320 grams of sulfur trioxide are required to produce 4.00 mol of sulfuric acid.