the question is below can anyone please help solve this step by step thanks?

[匿名]

2016-03-22 5:09 am

Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, the reaction of 5.0 grams of O2 with 6.0 grams of S can theoretically produce how many grams S O3 in this experiment?
S (s) + O2 (g) → SO3 (g) (not balanced)

回答 (1)

戇戇居士

2016-03-22 7:05 am

✔ 最佳答案

Molar mass of S = 32.06 g/mol
Molar mass of O₂ = 16.00 × 2 = 32.00 g/mol
Molar mass of SO₃ = 32.06 + 16.00 × 3 = 80.06 g/mol

2S(s) + 3O₂(g) → 2SO₃(g)
Theoretically, 2 moles of S reacts with 3 moles of O₂ to produce 2 moles of SO₃.

Initial number of moles of S = 6.0/32.06 = 0.1871 mol
Initial number of moles of O₂ = 5.0/32.00 = 0.1563 mol

If S completely reacts :
Number of moles of O₂ needed = 0.1871 × (3/2) = 0.2807 mol > 0.1563 mol
Hence, O₂ is the limiting reactant (completely reacts).

Number of moles of O₂ reacts = 0.1563 mol
Number of moles of SO₃ produces theoretically = 0.1563 × (2/3) = 0.1042 mol
Mass of SO₃ produces theoretically = 80.06 × 0.1042 = 8.34 g

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