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anonymous
 one year ago
A 0.250g sample of a magnesiumaluminum 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?
anonymous
 one year ago
A 0.250g sample of a magnesiumaluminum 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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sf Mg~(s)~+2HCl~(aq)~\rightarrow~ MgCl_2~(aq)~+~H_2~(g)\]\[\sf 2Al~(s) ~+~ 6HCl~(aq) ~\rightarrow~ 2AlCl_3~(s)~+~3H_2~(g)\] \[\sf x~g~Mg~\times~\frac{1~mol~Mg}{24.31~g~Mg} ~\times~\frac{1~mol~H_2}{1~mol~Mg}=~0.04114x~mol~H_2\]\[\sf y~g~Al ~\times ~\frac{1~mol~Al}{26.98~g~Al}~\times~ \frac{3~mol~H_2}{2~mol~Al}=~0.05560y~mol~H_2\]\[\sf x+y=0.250\]\[\sf n_{H_2} = 0.04114x+0.05560y=0.04114(0.250y)+0.05560y=0.0103+0.0145y\] \[\sf P_{H_2}V_{H_2}=n_{H_2}RT_{H_2}\]\[\sf P_{T} = P_{H_2O} +P_{H_2} \implies P_{H_2} = \frac{752~torr}{760~torr/atm}  \frac{30.0~torr}{760~torr/atm} = 0.0950~atm \]\[\sf n_{H_2}=\frac{PV}{RT} = \frac{(0.0950~atm \cdot 0.319~L)}{\left(\dfrac{0.08206~L\cdot atm}{mol\cdot K}\cdot (29+273)K\right)} = 0.0122~mol\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So then I guess I take the \(\sf n_{H_2}\) and I plug it back into my formula to find the total moles of \(\sf H_2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sf n_{H_2} = 0.0103+0.0145y\]\[\sf 0.0122 =0.0103+ 0.0145y\] \[\sf y=0.13\]\[\sf n_{H_2} = 0.0103 + 0.0145(0.13) =0.012~mol~H_2\] \[\sf 0.012 ~mol~H_2~\times~\frac{2~mol~Al}{3~mol~H_2}~\times~\frac{26.98~g~Al}{1~mol~Al} =0.22~g~Al\]\[\sf \boxed{\color{red}{\%~Al = \frac{0.22~g~Al}{0.250~g~comp}~\times ~100\% = 0.0088\%}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't think this is right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, the answer : \[\sf \%~Al = 53.1 ~\%\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, my error has to do with the equation of \(\sf n\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442107358544:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You know the answer because you gave up?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i didnt give up, I was reading through my notes haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Tbh i don't care about the answer, I just wanted to know if my steps were correct, and theyre not xD

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0@Jhannybean nicely done

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sf y~g~Al ~\times ~\frac{1~mol~Al}{26.98~g~Al}~\times~ \frac{3~mol~H_2}{2~mol~Al}=~0.05560y~mol~H_2\]\[\sf (0.05560(0.1310)~mol~H_2 = 0.00728~mol~H_2\]\[\sf 0.00728~mol~H_2 ~\times~\frac{2~mol~Al}{3~mol~H_2}~\times~\frac{26.98~g~Al}{1~mol~Al} =0.1309~g~Al\]\[\color{red}{\boxed{\sf \% Al= \frac{0.1309~g~Al}{0.250~g~comp}~\times~100\% = 52.3\%}}\]
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