Percentage of yield reaction question

Question

4. By bubbling $\ce{Cl2}$ in a solution containing $176.6~\mathrm{g}$ of $\ce{MgBr2}$ are obtained [sic!] $135~\mathrm{g}$ of $\ce{Br2}$. However, the yield of this reaction is not $100~\%$. How many grams of $\ce{Br2}$ t [sic!] are formed? What is the yield of the reaction? (MW [sic!] of $\ce{MgBr{2}}\ \text{[sic!]} = 184{,}13~\mathrm{amu}$; [sic!] MW [sic!] of $\ce{Br{2}}\ \text{[sic!]} = 159{,}82~\mathrm{amu}$ [sic!]). The reaction is:

$$\ce{MgBr2 + Cl2 -> MgCl2 + Br2}$$

I need help with this question please.

My thought was like this: $$176.6~\mathrm{g}/184.13=0.9591\ \mathrm{mol}\ \ce{MgBr2}$$ and this also equals the amount of $\ce{Br2}$ because it's the same ratio so $$0.9591\times159.82=153.28~\mathrm{g}\ \ce{Br2}$$

now the yield: $$135/153.28=0.88~\%$$ In my logic, the answer to how many grams are formed is $135~\mathrm{g}$ but it doesn't seem right.

Answer 1

I would expect a lower yield for Br2 because you are not just bubbling Cl2 in aqueous MgBr2! Once you start, there is also some now free Br2 available to react with the chlorine!

The neglected competing reaction is:

$\ce{Br2 + Cl2 ⇌ 2 BrCl}$

See this paper to get an idea of the affinity for the formation of the interhalogen compound.

But that is still only part of the story, because of a possible aqueous equilibrium reaction following the formation of any BrCl:

$\ce{BrCl(aq) + H2O ⇌ HOBr(aq) + H+ + Cl-}$

and even more active species can apparently be formed (see this paper).

Answer 2

As I already noted in the comments, you should always and consistently use correct units. And also, the typography of the question itself was terrible but that’s not something you can influence. Anyway to get to answering the question …

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I’m assuming that ‘grams of $\ce{MgBr2}$ t’ is supposed to mean something along the lines of theoretical yield. Also, I’m going to use $\mathrm{g/mol}$ as a unit for molar mass because that is what it should be. $\mathrm{amu}$ is a deprecated unit for atomic mass and cannot be used for molar mass without a conversion factor.

We thankfully are already given the reaction$$\ce{MgBr2 + Cl2 -> MgCl2 + Br2}\tag{1}$$ and it is thankfully the case that one mole of each reactant reacts to one mole of each product; i.e. all amounts of substance are theoretically equal. Thus, we can write the following:

$$\begin{align}n(\ce{MgBr2}) &= n (\ce{Br2})\tag{2.1}\\[0.4em] M = \frac mn\tag{3} &\Longrightarrow n = \frac mM\\[0.6em] \frac{m(\ce{MgBr2})}{M(\ce{MgBr2})} &= \frac{m(\ce{Br2})}{M(\ce{Br2})}\tag{2.2}\\[0.6em] m(\ce{Br2}) &= \frac{m (\ce{MgBr2}) \times M(\ce{Br2})}{M(\ce{MgBr2})}\tag{2.3}\\[0.6em] m(\ce{Br2}) &= \frac{176.6~\mathrm{g} \times 159.82~\mathrm{g/mol}}{184.13~\mathrm{g/mol}}\tag{2.4}\\[0.4em] &= 153.28~\mathrm{g}\tag{2.5}\end{align}$$

Therefore, your calculation luckily turned out to be correct in spite of not using units.

The question also asks for a yield. Yields are typically given in percent obtained of the theoretical maximum. In the lab, I typically calculate yields from substance amounts but here we can use mass because we already have the data. When calculating a percentage, you need to remember to multiply by $100$ before adding the $\%$ sign — which could be interpreted as a shorthand for $1/100$. Thus:

$$\frac{135~\mathrm{g}}{153.28~\mathrm{g}}= 0.88 = 88~\%\tag{4}$$

The units should, naturally, always cancel out in yields.

I mentioned before that the question is ill-phrased. In a strict literal sense, the answer to ‘How many grams of $\ce{Br2}$ t are formed?’ — without any information on what the random ‘t’ is supposed to be — would indeed read $135~\mathrm{g}$ but unless you are in a quiz show that is unlikely to be the ‘correct’ answer.