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In an experiment lanthanum $\ce{^{57}La}$ was reacted with $\ce{H2}$ to produce the non stoichiometric compound $\ce{LaH_{2.90}}$. Assuming that the compound contains $\ce{La^{2+}}$, $\ce{La^{3+}}$ and hydride ions, what is the % of $\ce{La^{3+}}$ present in $\ce{LaH_{2.90}}$ ?

I tried to find the percentage but I got a quadratic equation which has imaginary roots. Any help would be appreciated. I am aware of how to find the percentage of an element in a compound but it gets kind of fuzzy when we get to non stoichiometric compounds but I can understand the basics.

Here is what I tried:

Let the number of $\ce{La^2+}$ ions in $\ce{LaH_{2.90}}$ be x then the number of $\ce{La^3+}$ ions should be (2.9-x)^2 considering that the number of hydride ions are equal to the total number of La ions be it 2+ or 3+. Hence the quadratic expression emerges that x+(2.9-x)^2 = 1 which has no roots. I may be wrong so please do tell me if my assumption is incorrect.

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  • $\begingroup$ What compound would $\ce{La^{2+}}$ form? What about $\ce{La^{3+}}$? $\endgroup$
    – Buck Thorn
    Commented Nov 11, 2022 at 16:58
  • $\begingroup$ @BuckThorn the question should be read as LaH2.90 where 2.9 is in the subscript of H $\endgroup$ Commented Nov 11, 2022 at 17:01
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    $\begingroup$ I think you make this more complicated than need be. You can't have LaH2.90 in practice (only as a stoichiometric formula) so it has to be a mixture of LaH2 and LaH3. $\endgroup$
    – Buck Thorn
    Commented Nov 11, 2022 at 17:03
  • $\begingroup$ @BuckThorn well when we come to practicality all I can say is it wasn't my idea.. it was given by my teacher from general chemistry by Zumdahl & Decoste $\endgroup$ Commented Nov 11, 2022 at 17:05
  • $\begingroup$ dl.iranchembook.ir/ebook/General-Chemistry-604.pdf Q187 chapter 3. $\endgroup$ Commented Nov 11, 2022 at 17:05

3 Answers 3

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In an imaginary scenario where we could dissociate:

$$\ce{LaH_{2.9}->xLa^{+2} +yLa^{+3}}+2.9H^{-}$$

In terms of charge neutrality, we would need to satisfy:

$$2x+3y-2.9=0$$

In terms of stoichiometry, we would need to satisfy:

$$x+y=1$$

So we would get the following system of equations:

$$2x+3y=2.9$$

$$x+y=1$$

Solving it:

$$x=0.1$$

$$y=0.9$$

So there would be 10% of $\ce{La^{+2}}$ and 90% of $\ce{La^{+3}}$ out of all the $\ce{La}$ available.

The resulting dissociation reaction would be:

$$\ce{LaH_{2.9}->0.1La^{+2} +0.9La^{+3}}+2.9H^{-}$$

The total amount of species would be:

$$n=0.1+0.9+2.9=3.9$$

So the composition of each species in $\ce{LaH_{2.9}}$ would be:

$$X_{\ce{La^{+2}}}=\frac{0.1}{3.9}=0.02564=2.564\%$$

$$X_{\ce{La^{+3}}}=\frac{0.9}{3.9}=0.2308=23.08\%$$

$$X_{\ce{H^{-}}}=\frac{2.9}{3.9}=0.7436=74.36\%$$

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  • $\begingroup$ These percentages are probably not what the textbook wanted. They probably want to know what percentage of the metal has one or the other charge. $\endgroup$
    – Karsten
    Commented Nov 11, 2022 at 21:08
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    $\begingroup$ @Karsten The wording of the problem is not clear enough to conclude that, but either way, both have been answered. $\endgroup$
    – Sam202
    Commented Nov 11, 2022 at 21:11
  • $\begingroup$ The odd-numbered questions have the numerical answers (i.e. without the work) in the appendix. $\endgroup$
    – Karsten
    Commented Nov 11, 2022 at 23:01
  • $\begingroup$ You're correct. However, either OP misrepresented the problem in the textbook or someone who made an edit did, since the question posted here adds "present in $\ce{LaH_{2.9}}$" at the end of the statement, which is not said in the textbook statement. $\endgroup$
    – Sam202
    Commented Nov 11, 2022 at 23:05
  • $\begingroup$ You are also correct. This is the full wording in the document the OP referenced in the comments: "Lanthanum was reacted with hydrogen in a given experiment to produce the nonstoichiometric compound LaH2.90. Assuming that the compound contains $\ce{H2, La^2+, and La^3+}$, calculate the fractions of $\ce{La^2+ and La^3+}$ present." $\endgroup$
    – Karsten
    Commented Nov 11, 2022 at 23:18
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[OP] Let the number of $\ce{La^2+}$ ions in $\ce{LaH_{2.90}}$ be x then the number of $\ce{La^3+}$ ions should be (2.9-x)^2 considering that the number of hydride ions are equal to the total number of La ions be it 2+ or 3+. Hence the quadratic expression emerges that x+(2.9-x)^2 = 1 which has no roots.

The charge of the lanthanum ions is equal to the charge of the hydride ions. Your statement that the number of lanthanum ions is equal to that of hydride ions is incorrect.

Starting with the charges is more difficult, but possible. Let $c$ be the charge of the +2 ions in the formula unit. $2.9 - c$ will be the charge of the +3 ions. I can get the stoichiometric coefficient of the separate ions by dividing by 2 or 3, and these have to add up to one:

$$\frac{c}{2} + \frac{2.9 - c}{3} = 1$$

Multiplying by 6:

$$3c + 2(2.9 - c) = 6$$

Solving for c:

$$ c = 6 - 5.8 = 0.2$$

So $\ce{La^2+}$ contributes 0.2 of the charge, and $\ce{La^3+}$ 2.7 of the charge.

Of course, calling $x$ the stoichiometric coeffient of the $\ce{La^2+}$ ions gives an easier derivation, without fractions. You get

$$ x * 2 + (1 - x) * 3 = 2.9 $$

which gives $x = 0.1$.

Here are three ways to write the result:

$$0.1 \ce{La^2+ + 0.9 La^3+ + 2.9 H-}$$

or

$$\ce{\overset{+2}{La} _{0.1}\overset{+3}{La}_{0.9}H_{2.9}}$$

or

$$\ce{LaH2.9LaH3}$$

(with thanks to BuckThorn).

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  • $\begingroup$ Thanks for clarifying about the expression $\endgroup$ Commented Nov 13, 2022 at 5:46
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If you take $10$ $\ce{LaH2}$ and $90$ $\ce{LaH3}$ molecules, it makes a total of $100$ $\ce{La}$ atoms. But the number of $\ce{H}$ atoms is equal to $10·2 + 90·3 = 290$. This corresponds exactly to the formula $\ce{LaH_{2.90}}$

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