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    Just looking at an exam question and I'm a little confused however I'm confinced I'm nearly there.

    Original Equation: Na2CO3(s) + 2HNO3(aq) --> 2NaNO3(aq) + CO2(g) + H2O(l)

    The question asks what would the effect be if the experiment was repeated using 0.200mol dm-3 Sulphuric Acid (H2SO4), instead of 0.200mol dm-3 of Nitric Acid (HNO3).

    a) Write an equation for the new reaction:

    I have already done this:

    Na2CO3(s) + H2SO4(aq) --> Na2SO4(aq) + CO2(g) + H2O(g)

    b) What effects, if any, would using H2SO4 instead of HNO3 have on the volume of gas collected. Explain your answer. [2 Marks]

    I am guessing this has something to so with stoichiometry as both acids have the same moles and volume however in the 1st equation you are using (2)HNO3 and in the 2nd your using (1)H2SO4 so I am guessing that the gas collected is going to either double of half, but I have no idea which? :woo:

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    Hm I don't think it'd have any effect since the CO2 is the only gas formed which comes from the Na2CO3 therefore it'd be the same amount... no?

    I could be completely wrong though...
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    (Original post by Sbarkeri)
    Just looking at an exam question and I'm a little confused however I'm confinced I'm nearly there.

    Original Equation: Na2CO3(s) + 2HNO3(aq) --> 2NaNO3(aq) + CO2(g) + H2O(l)

    The question asks what would the effect be if the experiment was repeated using 0.200mol dm-3 Sulphuric Acid (H2SO4), instead of 0.200mol dm-3 of Nitric Acid (HNO3).

    a) Write an equation for the new reaction:

    I have already done this:

    Na2CO3(s) + H2SO4(aq) --> Na2SO4(aq) + CO2(g) + H2O(g)

    b) What effects, if any, would using H2SO4 instead of HNO3 have on the volume of gas collected. Explain your answer. [2 Marks]

    I am guessing this has something to so with stoichiometry as both acids have the same moles and volume however in the 1st equation you are using (2)HNO3 and in the 2nd your using (1)H2SO4 so I am guessing that the gas collected is going to either double of half, but I have no idea which? :woo:

    Cheers
    If the acid is the limiting reagent then there will be twice as much CO2, but if the sodium carbonate is the limiting reagent there is no difference except in the rate of the reaction.
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    Oh yes sorry forgot to mension but earlier on in the question it said "70cm3 of 0.200 mol dm-3 of HNO3 was added to 0.318g of Na2CO3 which is in excess. So as the sodium carbonate is in excess the nitric acid is the limiting reagent thus I should get 2x the amount of CO2?
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    (Original post by Sbarkeri)
    Oh yes sorry forgot to mension but earlier on in the question it said "70cm3 of 0.200 mol dm-3 of HNO3 was added to 0.318g of Na2CO3 which is in excess. So as the sodium carbonate is in excess the nitric acid is the limiting reagent thus I should get 2x the amount of CO2?
    Always important to let the dog see the rabbit...

    So:

    0.07 x 0.2 moles of HNO3 = 0.014 mol
    0.318/106 moles of Na2CO3 = 0.003132 mol

    in the reaction 1 mole Na2CO3 reacts with 2 moles of HNO3

    Therefore 0.003132 mol Na2CO3 reacts with 0.0063 mol HNO3

    However, you have 0.014 mol HNO3 so the ACID is in excess... !

    Unless of course you have made a typo and meant to write 0.318 MOLES of Na2CO3
 
 
 
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