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3 years ago

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

Chemistry

1 answer:

zalisa [80]3 years ago

3 0

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

<u><em>320 grams of sulfur trioxide are required to produce 4.00 mol of sulfuric acid.</em></u>

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