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Questions 8 and 9 refer to the analysis of a salt whose formula is $\ce{Pt(NH3)_xCl_y}$.

  1. $0.02$ moles [sic!] of this salt required $40.0~\mathrm{cm^3}$ of $2.0~\mathrm{mol\,l^{-1}}$ nitric acid for exact neutralisation.

    The number of moles of $\ce{NH3}$ per mole of salt is

    A 2

    B 4

    C 6

    D 8

  2. $0.02$ moles [sic!] of the salt were dissolved in nitric acid and excess silver(I) nitrate solution was added. The precipitate formed was filtered, washed and dried. It weighed $5.74~\mathrm{g}$.

    The number of moles of chloride per mole of the salt is

    A 1

    B 2

    C 3

    D 4

For question 8, I first found the no. of moles nitric acid to be $0.08\ \mathrm{mol}$, which is four times the number of moles of the salt. However, I just can't see how I would get $x$ from this. (The answer to this question is somehow B which makes even less sense since that's the number of moles of nitric acid for each mole of salt.)

For question 9, I tried to write a balanced equation for the formation of the precipitate but I don't think that's possible seeing that we don't know $x$ and $y$. Now, I know that the number of moles of the precipitate (silver chloride) is $0.04\ \mathrm{mol}$ which is twice than that of the salt. Yet again, I don't know how I would get $y$ from this. (The answer to this is also B, which is also the number of moles of precipitate for each mole of salt.) It would be appreciated if someone could explain how one can obtain the values lf $x$ and $y$ here.

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You're almost there!

For 8: For every mole of acid added, it reacts with a mole of the base included in the Pt(NH3)xCly, which is the ammonia. So you can figure out how many moles of ammonia have reacted, and compare this with the moles of salt you started with.

For 9: For every mole of AgCl you have formed, you have used up one mole of Cl from the Pt(NH3)xCly. So you can figure out how many moles of Cl you have captured, and compare this with the moles of salt you started with.

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