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

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

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Xavia 4 years 2021-09-02T06:21:42+00:00 1 Answers 176 views 0

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  1. Ladonna
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    2021-09-02T06:23:32+00:00
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    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.

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