Ask Question
28 June, 08:26

32 g of sulfur will react with 48 g of oxygen to produce 80 g of sulfur trioxide. if 20 g of sulfur and 100 g of oxygen are placed into a sealed container and allowed to react, what is the mass of sulfur trioxide in the container after the reaction is completed?

+4
Answers (1)
  1. 28 June, 11:44
    0
    The balanced chemical equation for the reaction of sulfur and oxygen to form sulfur trioxide is as follows:

    2S+3O_{2}/rightarrow 2SO_{3}

    That means 2 moles of sulfur reacts with 3 moles of oxygen to give 2 moles of sulfur trioxide.

    The reaction is always measured in terms of number of moles.

    First check for the limiting reagent here, the mole ratio is 1 S:1 SO_{3} and 1.5O_{2}:1SO_{3}

    Here, oxygen is in excess thus, sulfur is limiting reactant.

    Now, here, 20 g of sulfur and 100 g of oxygen is placed in the container. First convert the mass into moles as follows:

    For sulfur: molar mass is 32 g/mol

    Thus,

    n=/frac{m}{M}=/frac{20 g}{32 g}=0.625 mol

    Now, 1 mol of S gives 1 mol of SO_{3} thus, 0.625 mol of S gives 0.625 mol of SO_{3}.

    calculate mass of SO_{3} produced from molar mass and number of moles as follows:

    m=n*M=0.625 mol*80 g/mol=50 g

    Thus, mass of sulfur trioxide produced will be 50 g.
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “32 g of sulfur will react with 48 g of oxygen to produce 80 g of sulfur trioxide. if 20 g of sulfur and 100 g of oxygen are placed into a ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers