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    In an experiment, 0.10 mol of magnesium nitrate was heated. What is the maximum volume of gas, measured in dm3 at room temperature and pressure, could be obtained?
    2Mg(NO3)2(s)>>>>> 2MgO(s) + 4NO2(g) + O2(g)
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    Is that the full question could you please post the exam paper link?
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    (Original post by Neethan22)
    In an experiment, 0.10 mol of magnesium nitrate was heated. What is the maximum volume of gas, measured in dm3 at room temperature and pressure, could be obtained?
    2Mg(NO3)2(s)>>>>> 2MgO(s) + 4NO2(g) + O2(g)
    I got 6dm3 as the answer ( I don't know if this is right). This is how I got the answer. The ratios of magnesium nitrate:nitrogen oxide and oxygen is 2:5. 0.1 x 5/2 moles of gases are formed. This gives a total of 0.25 moles of gases. Therefore 0.25 x 24dm3 gives 6dm3 of gases.
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    (Original post by Neethan22)
    In an experiment, 0.10 mol of magnesium nitrate was heated. What is the maximum volume of gas, measured in dm3 at room temperature and pressure, could be obtained?
    2Mg(NO3)2(s)>>>>> 2MgO(s) + 4NO2(g) + O2(g)
    IS it 6.0 dm3 ?

    if it is then

    there is 0.1 mol of Mg(NO3)2 , there is two gases obtained, NO2 and O2, the mole ratio between Mg(NO3)2 and NO2 is 1:2 and the mole ratio between Mg(NO3)2 and O2 is 2:1

    Therefore the number of moles of NO2 is 0.1*2 = 0.2 mol
    And number of moles of O2 is 0.1/2 = 0.05 mol

    The gases are at room temperature and pressure so number of moles = volume/24

    Volume of NO2 produced is 0.2*24 = 4.8 dm3
    Volume of O2 produced is 0.05*24 = 1.2 dm3
    So total volume of gases produced is 4.8 + 1.2 = 6.0 dm3
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    (Original post by Sunethra)
    I got 6dm3 as the answer ( I don't know if this is right). This is how I got the answer. The ratios of magnesium nitrate:nitrogen oxide and oxygen is 2:5. 0.1 x 5/2 moles of gases are formed. This gives a total of 0.25 moles of gases. Therefore 0.25 x 24dm3 gives 6dm3 of gases.
    ya it's 6dm3
    thanks for the explanation
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    (Original post by PlayerBB)
    IS it 6.0 dm3 ?

    if it is then

    there is 0.1 mol of Mg(NO3)2 , there is two gases obtained, NO2 and O2, the mole ratio between Mg(NO3)2 and NO2 is 1:2 and the mole ratio between Mg(NO3)2 and O2 is 2:1

    Therefore the number of moles of NO2 is 0.1*2 = 0.2 mol
    And number of moles of O2 is 0.1/2 = 0.05 mol

    The gases are at room temperature and pressure so number of moles = volume/24

    Volume of NO2 produced is 0.2*24 = 4.8 dm3
    Volume of O2 produced is 0.05*24 = 1.2 dm3
    So total volume of gases produced is 4.8 + 1.2 = 6.0 dm3
    yes it's 6dm3
    thanks a lot for your explanation
 
 
 
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