A 0.250-g sample of… - QuestionCove
OpenStudy (jhannybean):

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?

2 years ago
OpenStudy (jhannybean):

$\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 years ago
OpenStudy (jhannybean):

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 $$\sf H_2$$

2 years ago
OpenStudy (jhannybean):

$\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\%}}$

2 years ago
OpenStudy (jhannybean):

I don't think this is right.

2 years ago
OpenStudy (jhannybean):

Oh, the answer : $\sf \%~Al = 53.1 ~\%$

2 years ago
OpenStudy (jhannybean):

Ok, my error has to do with the equation of $$\sf n$$

2 years ago
OpenStudy (anonymous):

Attach files.

2 years ago
OpenStudy (jhannybean):

|dw:1442107358544:dw|

2 years ago
OpenStudy (anonymous):

You know the answer because you gave up?

2 years ago
OpenStudy (jhannybean):

i didnt give up, I was reading through my notes haha

2 years ago
OpenStudy (jhannybean):

Tbh i don't care about the answer, I just wanted to know if my steps were correct, and theyre not xD

2 years ago
OpenStudy (jhannybean):

I figured it out

2 years ago
OpenStudy (photon336):

@Jhannybean nicely done

2 years ago
OpenStudy (jhannybean):

$\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\%}}$

2 years ago
OpenStudy (jhannybean):

and ty @Photon336

2 years ago