Consider the following reaction: 2SO2(g) + O2(g) ->> 2SO3(g)?

[匿名]

2017-10-25 12:48 pm

if 285.3 mL of SO2 is allowed to react with 158.9 mL of 02(both measured at 315 K and 50 mmHg) what is the limiting reactant and theoretical yield of SO3 in moles? If 187.2 mL of SO3 is collected measured at 315K and 50 mm Hg, what is the percent yield for the reaction?

回答 (1)

wanszeto

2017-10-25 1:02 pm

✔ 最佳答案

At constant pressure and temperature, mole ratio of gas is equal to volume ratio.

2SO₂(g) + O₂(g) → 2SO₃(g)
Volume ratio in reaction SO₂ : O₂ = 2 : 1

When 285.3 mL SO₂ completely reacted, O₂ needed = (285.3 mL) × (1/2) = 142.65 mL < 158.9 mL
Hence, O₂ is in excess, and SO₂ is the limiting reactant/reagent.

According to the above equation, volume ratio in reaction SO₂ : SO₃ = 2 : 2 = 1 : 1
Maximum number of moles of SO₃ produced = Initial volume of SO₂ = 285.3 mL
Percent yield of the reaction = (187.2/283.5) × 100% = 65.6%

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