CH,(9)+20,(9) –2 H,0(g)+CO,(9) Use this chemical equation to answer the questions in the table below. Suppose 80.0 mmol of CH and O None. 160. mmol of O, are added to an O Some, but less than 80.0 mmol. empty flask. How much CH will be O 80.0 mmol. in the flask at equilibrium? O More than 80.0 mmol. Suppose 180. mmol of H,O and O None. 90.0 mmol of CO, are added to an O Some, but less than 180. mmol. empty flask. How much H,0 will be O 180. mmol. in the flask at equilibrium? O More than 180. mmol.

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Methane and oxygen react to form water and carbon dioxide, like this:
CH,(9)+20,(9) → 2 H,O(g)+CO,(9)
Use this chemical equation to answer the questions in the table below.
Suppose 80.0 mmol of CH, and
O None.
160. mmol of O, are added to an
O Some, but less than 80.0 mmol.
empty flask. How much CH, will be
O 80.0 mmol.
in the flask at equilibrium?
O More than 80.0 mmol.
Suppose 180. mmol of H, O and
O None.
90.0 mmol of CO, are added to an
O Some, but less than 180. mmol.
empty flask. How much H,O will be
O 180. mmol.
in the flask at equilibrium?
O More than 180. mmol.
Transcribed Image Text:Methane and oxygen react to form water and carbon dioxide, like this: CH,(9)+20,(9) → 2 H,O(g)+CO,(9) Use this chemical equation to answer the questions in the table below. Suppose 80.0 mmol of CH, and O None. 160. mmol of O, are added to an O Some, but less than 80.0 mmol. empty flask. How much CH, will be O 80.0 mmol. in the flask at equilibrium? O More than 80.0 mmol. Suppose 180. mmol of H, O and O None. 90.0 mmol of CO, are added to an O Some, but less than 180. mmol. empty flask. How much H,O will be O 180. mmol. in the flask at equilibrium? O More than 180. mmol.
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