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  1. 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.

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    $\begingroup$ Welcome to Chemistry! This is a homework question. We have a policy which states that ‎you should show your thoughts and/or efforts into solving the problem. It'll make us certain that ‎we aren't doing your homework for you. Otherwise, this question may get closed.‎ Please edit in your full reasoning or thoughts on this. $\endgroup$ – M.A.R. ಠ_ಠ Jan 14 '17 at 18:13
  • $\begingroup$ ok, just did :) sorry i didnt know $\endgroup$ – sapir Jan 14 '17 at 19:58
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    $\begingroup$ You should always and consistently use correct units in your equations. $\endgroup$ – Jan Jan 14 '17 at 20:55
  • $\begingroup$ ok... but my answers are right? $\endgroup$ – sapir Jan 14 '17 at 23:31
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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 …


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.

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  • $\begingroup$ Thank you very much, i appreciate your help. Im study in Italy in english, this is why the questions are so ill phrased. $\endgroup$ – sapir Jan 16 '17 at 15:53
  • $\begingroup$ @sapir Rather than just saying ‘thanks’ you can also accept this answer if it helped you by clicking the green checkmark. And once you have 15 reputation, you can also vote it up. $\endgroup$ – Jan Jan 16 '17 at 21:51

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