Methane and oxygen react to form water and carbon dioxide, like this:CH4(g)+2O2(g)→2H2O(g)+CO2(g)Use this chemical equation to answer the questions in the table below.Suppose 165.mmol of CH4 and 330.mmol of O2 are added to an empty flask. How much CH4 will be in the flask at equilibrium? None.Some, but less than 165.mmol.165.mmol.More than 165.mmol.Suppose 70.0mmol of H2O and 35.0mmol of CO2 are added to an empty flask. How much CH4 will be in the flask at equilibrium?
Question
Methane and oxygen react to form water and carbon dioxide, like this:CH4(g)+2O2(g)→2H2O(g)+CO2(g)Use this chemical equation to answer the questions in the table below.Suppose 165.mmol of CH4 and 330.mmol of O2 are added to an empty flask. How much CH4 will be in the flask at equilibrium? None.Some, but less than 165.mmol.165.mmol.More than 165.mmol.Suppose 70.0mmol of H2O and 35.0mmol of CO2 are added to an empty flask. How much CH4 will be in the flask at equilibrium?
Solution
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The reaction CH4(g) + 2O2(g) → 2H2O(g) + CO2(g) is a complete reaction where all the reactants are used up to form the products. Therefore, if 165 mmol of CH4 and 330 mmol of O2 are added to an empty flask, all of the CH4 will react with the O2 to form H2O and CO2. So, at equilibrium, there will be no CH4 left in the flask. The answer is "None".
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If 70.0 mmol of H2O and 35.0 mmol of CO2 are added to an empty flask, there will be no CH4 in the flask at equilibrium. This is because the reaction has already occurred to form the products (H2O and CO2), and there are no reactants (CH4 and O2) present to reverse the reaction and form CH4. The answer is "None".
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