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Jhannybean

  • one year ago

A 0.250-g sample of a magnesium-aluminum alloy dissolves completely in an excess of HCl (aq). When the liberated \(\sf H_2~(g)\) is collected over water at 29 ∘C and 752 torr, the volume is found to be 319 mL . The vapor pressure of water at 29 ∘C is 30.0 torr. What is the mass percentage of aluminum in this alloy?

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  1. Jhannybean
    • one year ago
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    \[\sf Mg~(s)~+2HCl~(aq)~\rightarrow~ MgCl_2~(aq)~+~H_2~(g)\]\[\sf 2Al~(s) ~+~ 6HCl~(aq) ~\rightarrow~ 2AlCl_3~(s)~+~3H_2~(g)\] \[\sf x~g~Mg~\times~\frac{1~mol~Mg}{24.31~g~Mg} ~\times~\frac{1~mol~H_2}{1~mol~Mg}=~0.04114x~mol~H_2\]\[\sf y~g~Al ~\times ~\frac{1~mol~Al}{26.98~g~Al}~\times~ \frac{3~mol~H_2}{2~mol~Al}=~0.05560y~mol~H_2\]\[\sf x+y=0.250\]\[\sf n_{H_2} = 0.04114x+0.05560y=0.04114(0.250-y)+0.05560y=0.0103+0.0145y\] \[\sf P_{H_2}V_{H_2}=n_{H_2}RT_{H_2}\]\[\sf P_{T} = P_{H_2O} +P_{H_2} \implies P_{H_2} = \frac{752~torr}{760~torr/atm} - \frac{30.0~torr}{760~torr/atm} = 0.0950~atm \]\[\sf n_{H_2}=\frac{PV}{RT} = \frac{(0.0950~atm \cdot 0.319~L)}{\left(\dfrac{0.08206~L\cdot atm}{mol\cdot K}\cdot (29+273)K\right)} = 0.0122~mol\]

  2. Jhannybean
    • one year ago
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    So then I guess I take the \(\sf n_{H_2}\) and I plug it back into my formula to find the total moles of \(\sf H_2\)

  3. Jhannybean
    • one year ago
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    \[\sf n_{H_2} = 0.0103+0.0145y\]\[\sf 0.0122 =0.0103+ 0.0145y\] \[\sf y=0.13\]\[\sf n_{H_2} = 0.0103 + 0.0145(0.13) =0.012~mol~H_2\] \[\sf 0.012 ~mol~H_2~\times~\frac{2~mol~Al}{3~mol~H_2}~\times~\frac{26.98~g~Al}{1~mol~Al} =0.22~g~Al\]\[\sf \boxed{\color{red}{\%~Al = \frac{0.22~g~Al}{0.250~g~comp}~\times ~100\% = 0.0088\%}}\]

  4. Jhannybean
    • one year ago
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    I don't think this is right.

  5. Jhannybean
    • one year ago
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    Oh, the answer : \[\sf \%~Al = 53.1 ~\%\]

  6. Jhannybean
    • one year ago
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    Ok, my error has to do with the equation of \(\sf n\)

  7. anonymous
    • one year ago
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    Attach files.

  8. Jhannybean
    • one year ago
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    |dw:1442107358544:dw|

  9. anonymous
    • one year ago
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    You know the answer because you gave up?

  10. Jhannybean
    • one year ago
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    i didnt give up, I was reading through my notes haha

  11. Jhannybean
    • one year ago
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    Tbh i don't care about the answer, I just wanted to know if my steps were correct, and theyre not xD

  12. Jhannybean
    • one year ago
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    I figured it out

  13. Photon336
    • one year ago
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    @Jhannybean nicely done

  14. Jhannybean
    • one year ago
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    \[\sf y~g~Al ~\times ~\frac{1~mol~Al}{26.98~g~Al}~\times~ \frac{3~mol~H_2}{2~mol~Al}=~0.05560y~mol~H_2\]\[\sf (0.05560(0.1310)~mol~H_2 = 0.00728~mol~H_2\]\[\sf 0.00728~mol~H_2 ~\times~\frac{2~mol~Al}{3~mol~H_2}~\times~\frac{26.98~g~Al}{1~mol~Al} =0.1309~g~Al\]\[\color{red}{\boxed{\sf \% Al= \frac{0.1309~g~Al}{0.250~g~comp}~\times~100\% = 52.3\%}}\]

  15. Jhannybean
    • one year ago
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    and ty @Photon336

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