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alphabeta

  • one year ago

Need help with a titration task (attached file). How should I best approach this problem?

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  1. alphabeta
    • one year ago
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  2. Rushwr
    • one year ago
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    First write the balanced chemical equation Since we are given with the concentration and volume of sulfuric apply C=n/V and find the no. of moles (n) According to the stoichiometric ratios find the no. of moles of ammonia. And then apply C=n/V and find the concentration (C)

  3. Rushwr
    • one year ago
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    H2SO4 + 2NH3 --> (NH4)2SO4 According to the equation we can see the no. of moles ammonia is twice the no. of moles of sulfuric . So find the no. of moles of sulfuric and multiply it by 2 to find the no. of moles of ammonia.

  4. alphabeta
    • one year ago
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    Thank you for your response @Rushwr! It sounds like what I first did, but still, I didn't manage to get the correct answer. Could you help me see where I'm wrong? How I calculated it: 2NH3 + H2SO4 --> (NH4)2SO4 n(H2SO4) = (0.150 mol dm^-3)/(0.0201 dm^3) = 7.46268... mol n(NH3) = 2*n(H2SO4) = 2*7.46268... = 14.92537... mol c(NH3) = nv = 14.92537... * 0.025 = 0.373 mol dm^-3 The mark scheme says that the concentration of NH3 is 0.241 mol dm^-3. Again; thank you for your response!

  5. Rushwr
    • one year ago
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    There's a calculation error in the moles of sulfuric.

  6. Rushwr
    • one year ago
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    Actually you don't have to convert the volumes to dm^3 cuz it cancels off in the final equation. Let me show you. C=n/V right? And we know no. of moles of ammonia is twice of sulfuric. No. of moles of sulfuric = n \[n= 0.15moldm ^{-3} * 20.1cm ^{3}\] no. of moles of ammonia is 2n right? So \[2n = 0.15moldm ^{-3}* 20.1cm ^{3} * 2\] Now apply the values for C=n/V \[C= \frac{ 0.15moldm ^{-3} * 20.1cm ^{3} * 2 }{ 25cm ^{3} }\] Since cm^3 is cancelled off we don't actually have to solve that. so C= 0.241moldm^-3

  7. alphabeta
    • one year ago
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    Ahhh, now I see! I was a little too fast when thinking about that c = n/v formula and did n = c/v instead. And now I realize that it was unnecessary to convert dm^3 to cm^3 as you said. THANK YOU SO MUCH!

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