Sulfur dioxide and oxygen react to form sulfur trioxide, like this: 2 SO,(9)+O,(9) → 2 SO3(9) Use this chemical equation to answer the questions in the table below. Suppose 155. mmol of SO, and None. 77.5 mmol of O, are added to an O Some, but less than 155. mmol. empty flask. How much SO, will be O 155. mmol. in the flask at equilibrium? O More than 155. mmol. O None. Suppose 55.0 mmol of SO, are O Some, but less than 55.0 mmol. added to an empty flask. How much SO, will be in the flask at O 55.0 mmol. equilibrium? O More than 55.0 mmol.

Chemistry: Principles and Reactions
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Chapter5: Gases
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Sulfur dioxide and oxygen react to form sulfur trioxide, like this:
2 SO,(9)+O,(9) → 2 SO3(9)
Use this chemical equation to answer the questions in the table below.
Suppose 155. mmol of SO, and
None.
77.5 mmol of O, are added to an
O Some, but less than 155. mmol.
empty flask. How much SO, will be
O 155. mmol.
in the flask at equilibrium?
O More than 155. mmol.
O None.
Suppose 55.0 mmol of SO, are
O Some, but less than 55.0 mmol.
added to an empty flask. How much
SO, will be in the flask at
O 55.0 mmol.
equilibrium?
O More than 55.0 mmol.
Transcribed Image Text:Sulfur dioxide and oxygen react to form sulfur trioxide, like this: 2 SO,(9)+O,(9) → 2 SO3(9) Use this chemical equation to answer the questions in the table below. Suppose 155. mmol of SO, and None. 77.5 mmol of O, are added to an O Some, but less than 155. mmol. empty flask. How much SO, will be O 155. mmol. in the flask at equilibrium? O More than 155. mmol. O None. Suppose 55.0 mmol of SO, are O Some, but less than 55.0 mmol. added to an empty flask. How much SO, will be in the flask at O 55.0 mmol. equilibrium? O More than 55.0 mmol.
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