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    i suck with titration cal plz help with this question
    Calculate (a) the moles and (b) the mass of magnesium carbonate at the start if 0.2 moles of sulfuric acid is added to the magnesium carbonate and the excess sulfuric acid made up to a 250 cm3 solution. 25 cm3 of this solution required 0.03 moles of sodium hydroxide for neutralisation.
    MgCO3 + H2SO4 → MgSO4 + H2O + CO2
    H2SO4 + 2 NaOH → NaCl + H2O
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    You'll do yourself no favours by asking people for the answer. I'd suggest going onto youtube and watching a few titration videos
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    (Original post by thad33)
    You'll do yourself no favours by asking people for the answer. I'd suggest going onto youtube and watching a few titration videos
    I have already done that still can't get the right answer

    the answer is 0.05g

    first i found the mol of Naoh- which is 7.5x10-4
    then h2s04- 3.75x10-4
    as 1/10 is used mol of h2so4 is 3.75x10-3
    mol of h2so4 used is 0.19625

    i took it away from the initial mol but i get the wrong answer

    BTW i always do the question myself before asking people
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    a) should be 0.05 mol and b) should be 4.125g

    You're already given the moles of 2NaOH, it's 0.03. So the moles of H2SO4 is 0.015 as the ratio is 2:1. Then you scale it up by 10 because 0.015 is the moles in 25cm^3, you want the moles in 250. So you have 0.15 moles of unreacted H2SO4, you started with 0.2 moles of it which means 0.05 must have reacted with MgCO3 as its a 1:1 reaction.

    So a) is 0.05

    Then you times that by the Mr of MgCO3 for the mass. 84.3 x 0.05 = 4.125


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    (Original post by yeah1106)
    i suck with titration cal plz help with this question
    Calculate (a) the moles and (b) the mass of magnesium carbonate at the start if 0.2 moles of sulfuric acid is added to the magnesium carbonate and the excess sulfuric acid made up to a 250 cm3 solution. 25 cm3 of this solution required 0.03 moles of sodium hydroxide for neutralisation.
    MgCO3 + H2SO4 → MgSO4 + H2O + CO2
    H2SO4 + 2 NaOH → NaCl + H2O
    Hi,
    I got the answer to be a)0.05 mol b) 4.22g
    Double check your markscheme to see if this correct first then I will post the method
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    (Original post by AshvinSeetul)
    Hi,
    I got the answer to be a)0.05 mol b) 4.22g
    Double check your markscheme to see if this correct first then I will post the method

    Its right
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    (Original post by yeah1106)
    Its right
    Ok first just think logically, there are two reactions going on here so treat them separately.

    1. Find the number of moles in the neutralisation reaction
    H2SO4 + 2NaOH.....etc
    The ratio is 1:2
    So we know the number of moles of NaOH which is 0.03mol, therefore as part of the ratio we need to divide 0.03 by 2 to get the number of moles of H2SO4= 0.015mol.
    However this is only in 25cm3 of acid as the question tells us the solution has been diluted to 250cm3. 25cm3------> 250cm3 you need to multiply the number of moles of H2SO4 by x10
    This will give you 0.15 mol of acid

    2. Find number of moles used in the first reaction
    MgSO4 + H2SO4....etc
    We are told that the first reaction is in excess( indirectly) which is done using 0.2 moles of acid and we know that 0.15moles which was the left over of the reaction was used to react with the base NaOH. Therefore 0.2-0.15= 0.05mol used in the reaction

    3. Find mass of MgSO4
    It's a 1:1 ratio so use the equation mass = molecular mass x number of moles

    Molecular mass is of MgSO4= 84.3
    Number of moles is 0.05mol

    Put in the equation and you answer is 4.22g (3 sf)

    Hope this helps!!!!
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    (Original post by AshvinSeetul)
    Ok first just think logically, there are two reactions going on here so treat them separately.

    1. Find the number of moles in the neutralisation reaction
    H2SO4 + 2NaOH.....etc
    The ratio is 1:2
    So we know the number of moles of NaOH which is 0.03mol, therefore as part of the ratio we need to divide 0.03 by 2 to get the number of moles of H2SO4= 0.015mol.
    However this is only in 25cm3 of acid as the question tells us the solution has been diluted to 250cm3. 25cm3------> 250cm3 you need to multiply the number of moles of H2SO4 by x10
    This will give you 0.15 mol of acid

    2. Find number of moles used in the first reaction
    MgSO4 + H2SO4....etc
    We are told that the first reaction is in excess( indirectly) which is done using 0.2 moles of acid and we know that 0.15moles which was the left over of the reaction was used to react with the base NaOH. Therefore 0.2-0.15= 0.05mol used in the reaction

    3. Find mass of MgSO4
    It's a 1:1 ratio so use the equation mass = molecular mass x number of moles

    Molecular mass is of MgSO4= 84.3
    Number of moles is 0.05mol

    Put in the equation and you answer is 4.22g (3 sf)

    Hope this helps!!!!
    Thanks so much turn out i read the question wrong i though it said 25cm3 of 0.03moldm-3 such as idiot i am
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    Ah yeah 4.215, not 4.125. I mixed the numbers around.

    I find it helps to write out and label every piece of information that you're given before even attempting the question


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