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  • Message 1. 

    Posted by U15226582 (U15226582) on Sunday, 28th April 2013

    Found another calculation I was stuck on in the credit paper 2007 question 20bi)&ii) thanks!

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  • Message 2

    , in reply to message 1.

    Posted by MissChemTeacher_Bitesize (U15172316) on Monday, 29th April 2013


    Cheers for your message. How are you feeling about the exam on Wed?

    I'll go through all of Q20 so hopefully you'll get the hang of it.

    20a) Using the data from the Q you've to calculate the number of moles of hydrochloric acid.

    n moles of HCl = concentration x volume in litres
    n moles of HCl = 4.0 x 50/1000 (to convert into litres)
    n moles of HCl = 0.2 moles

    bi) You now need to use the balanced equation to work out how many moles of calcium carbonate are needed to neutralise the acid.

    2HCl + CaCO₃ ----------> CaCl₂ + H₂O + CO₂

    You can see from the balanced equation that 2 moles of acid reacts with one mole of calcium carbonate (the big numbers in front of the different compounds)

    From Balanced Eq: 2HCl + CaCO₃
    Mole Ratio:................2.....:.......1
    From Q:....................0.2....:.....0.1

    We just worked out that we had 0.2 moles of acid so we'd need to half that to calculate the number of moles of calcium carbonate.

    bii) We now know how many moles of calcium carbonate we have. We can use the other triangle to calculate the mass of the carbonate.

    mass of CaCO₃ = n of moles x gfm
    mass of CaCO₃ = 0.1 x 100
    mass of CaCO₃ = 10g

    Hope this helps you out. If you have any more questions feel free to get in touch.

    All the best for Wednesday!

    Miss_Chem smiley - cool

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