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1) 1.00 L of methane, CH₄, reacts completely with excess oxygen to form carbon dioxide and water. What volumes of the two products are produced at this pressure and temperature?

A. 1.00 L CO₂ and 2.00 L H₂O
B. 1.00 L CO₂ and 4.00 L H₂O
C. 2.00 L CO₂ and 2.00 L H₂O
D. 2.00 L CO₂ and 4.00 L H₂O

Answer :

Final answer:

The reaction between methane and oxygen produces 1.00 L of carbon dioxide and 2.00 L of water at STP.


Explanation:

The balanced chemical equation for the reaction between methane (CH4) and oxygen (O2) is:



CH4 + 2O2 → CO2 + 2H2O



From the balanced equation, we can see that for every 1 mole of methane, we get 1 mole of carbon dioxide and 2 moles of water. Since we're given the volume of methane as 1.00 L, we can assume it is at standard temperature and pressure (STP). At STP, 1 mole of any ideal gas occupies a volume of 22.4 L.



Therefore, 1.00 L of methane at STP corresponds to 1.00/22.4 = 0.0446 moles of methane. Using the stoichiometry of the balanced equation, we can determine the moles of carbon dioxide and water produced:



1 mole of methane produces 1 mole of carbon dioxide and 2 moles of water



0.0446 moles of methane will produce 0.0446 moles of carbon dioxide and 2(0.0446) = 0.0892 moles of water



Since 1 mole of any ideal gas occupies a volume of 22.4 L at STP, we can convert the moles to liters:



0.0446 moles of carbon dioxide is approximately 0.0446(22.4) = 1.00 L



0.0892 moles of water is approximately 0.0892(22.4) = 2.00 L



Therefore, the correct answer is (A) 1.00 L CO2 and 2.00 L H2O.


Learn more about Stoichiometry here:

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