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    Calculate (a) the moles and (b) the mass of magnesium carbonate at the start if 0.200 moles of sulfuric acid is added
    to the magnesium carbonate and the excess sulfuric acid made up to a 250 cm3
    solution. 25.0 cm3
    of this solution
    required 0.0300 moles of sodium hydroxide for neutralisation.
    The answers are 0.05 mol and 4.22g but I don't know how to get there. Help Please!!
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    (Original post by PenguinNinja)
    Calculate (a) the moles and (b) the mass of magnesium carbonate at the start if 0.200 moles of sulfuric acid is added
    to the magnesium carbonate and the excess sulfuric acid made up to a 250 cm3
    solution. 25.0 cm3
    of this solution
    required 0.0300 moles of sodium hydroxide for neutralisation.
    The answers are 0.05 mol and 4.22g but I don't know how to get there. Help Please!!
    N = C X V / 1000
    (moles = concentration x volume) / 1000
    N = (0.200 x 250) / 1000
    N = 0.05

    N = M / Mr
    (moles = mass / molar mass)
    We know N is 0.05
    We know Mr of MgCO3 is 84
    So we have to rearrange the equation:
    M = N x Mr
    M = 0.05 x 84
    M = 4.2g

    (Mr of MgCO3 might be different depending on how many decimal places you're using - I just rounded it up to 84 for the sake of explaining)
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    (Original post by Petulia)
    N = C X V / 1000
    (moles = concentration x volume) / 1000
    N = (0.200 x 250) / 1000
    N = 0.05

    N = M / Mr
    (moles = mass / molar mass)
    We know N is 0.05
    We know Mr of MgCO3 is 84
    So we have to rearrange the equation:
    M = N x Mr
    M = 0.05 x 84
    M = 4.2g

    (Mr of MgCO3 might be different depending on how many decimal places you're using - I just rounded it up to 84 for the sake of explaining)
    Thanks so much! You made it really easy to understand.:cookie:
 
 
 

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