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    I'm doing the new a-level chemistry course and our teacher gave us few questions related to this new course and there is one question worth 1 mark, mark scheme says the answer is D(130 cm^3) but i have no idea why it is when the total volume is just 120cm^3.The question is attached to this post. If anyone can explain it i would appreciate it.
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    (Original post by Learner0)
    I'm doing the new a-level chemistry course and our teacher gave us few questions related to this new course and there is one question worth 1 mark, mark scheme says the answer is D(130 cm^3) but i have no idea why it is when the total volume is just 120cm^3.The question is attached to this post. If anyone can explain it i would appreciate it.
    The volume of the reactants is 120cm^3 yes, but that does not necessarily mean that the volume of gases will be 120cm^3 after the reaction has taken place.

    Provided pressure and temperature are constant, volume is proportional to the number of moles (using ideal gas equation, pV=nRT). Using this, you have to work out the limiting reagent. If there is 20cm^3 of ethane and 100cm^3 of oxygen, there must be 5 moles of O2 for every mole of ethane. However, looking at the stoichiometric coefficients (the numbers in the chemical equation) only 3.5 moles of O2 are needed for every mole of ethane - hence oxygen is the reagent in excess. This means that ethane is the limiting reagent. For every mole of ethane that combusts, two moles of CO2 and three moles of steam (since the temperature is greater than 100°C) are formed. So the total volume of gas produced is the sum of the volume of steam, carbon dioxide and the excess oxygen which was left over.

    Volume of CO2 = 2 x 20cm^3 = 40cm^3
    Volume of H20 (steam) = 3 x 20cm^3 = 60cm^3
    Volume of O2 left over = 100cm^3 - 3.5 x 20cm^3 = 30cm^3

    Hence the total volume of gas produced = 40 + 60 + 30 = 130cm^3.

    Hope that helps!
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    (Original post by Jpw1097)
    The volume of the reactants is 120cm^3 yes, but that does not necessarily mean that the volume of gases will be 120cm^3 after the reaction has taken place.

    Provided pressure and temperature are constant, volume is proportional to the number of moles (using ideal gas equation, pV=nRT). Using this, you have to work out the limiting reagent. If there is 20cm^3 of ethane and 100cm^3 of oxygen, there must be 5 moles of O2 for every mole of ethane. However, looking at the stoichiometric coefficients (the numbers in the chemical equation) only 3.5 moles of O2 are needed for every mole of ethane - hence oxygen is the reagent in excess. This means that ethane is the limiting reagent. For every mole of ethane that combusts, two moles of CO2 and three moles of steam (since the temperature is greater than 100°C) are formed. So the total volume of gas produced is the sum of the volume of steam, carbon dioxide and the excess oxygen which was left over.

    Volume of CO2 = 2 x 20cm^3 = 40cm^3
    Volume of H20 (steam) = 3 x 20cm^3 = 60cm^3
    Volume of O2 left over = 100cm^3 - 3.5 x 20cm^3 = 30cm^3

    Hence the total volume of gas produced = 40 + 60 + 30 = 130cm^3.

    Hope that helps!
    Thanks it totally make scene.
 
 
 
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