insomaniacal - 3-3-2010 at 13:05
Basically, I have a problem, and want to know the best way to do it. Thanks a ton to anyone who can help me out.
Basically, I have 5g of Aluminum, and want to create Aluminum Oxide, using all of the Aluminum. I calculated the amount of oxygen I would need by
using ratios. .185 moles of Aluminum * (3mol O/4 mol Al), and got .13875 as an answer. Now, converting this to Liters gives me 3.108. This is the same
answer as on the answer sheet (3.1L), so I'm assuming I did everything correctly. If not, please feel free to tell me what I did wrong.
Now, the next question is asking how many grams of aluminum oxide will be created, and where the confusion starts. The balanced chemical equation is
4Al + 302 --> 2Al2O3 . So, the ratio looks like it would be 4:6:4:6, however, doing this, I get a really unrealistic answer (over 102 grams, which
would be equal to one mole.). Just for laughs, I did it with the unbalanced equation (1:2:2:3), and got the correct numbers for the answer (94.35).
However,the real answer is 9.435. I have no idea how to get there using only this ratio method.
Basically, I had taken the .185 moles, multiplied it it's ratio (4), then did the same for oxygen (3), added them together, and multipled by the molar
mass of Al2O3, which is 102. I did the same method when using 1:2:2:3 .
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Now, I did finally find a way to get the correct answer, using limiting reactants, but I'm almost 100% sure we never learned about those. I'll admit I
haven't payed attention 100% of the time in this class, and I've been absent for at least 3 periods due to sickness.
Basically, since there is enough of each so that they both run out at the same time, I took the .185, divided it by 2, then multiplied that by the
molar mass (102), and got the correct answer. This also works if I took the .13875, multiplied it by two (since the formula says o2), divided by
three, and multiplied by the molar mass (102).
Is that the only way to do it, or am I missing something with the ratios? I've tried asking this around other corners of the internet, and have gotten
mostly flame, along with some unrealistic answers (like over 30g, which just by using logic seems impossible).
Thanks a lot to anyone who replies!
Magpie - 3-3-2010 at 14:50
The balanced chemical equation is 4Al + 302 --> 2Al2O3
You used 5g Al, so:
wt Al2O3 = [2(MW Al2O3)/4(MW Al)](5g)
Limiting reactant doesn't enter into this as you have a stoichiometric amount of both reactants.
Ramiel - 4-3-2010 at 01:35
Or the way I would have thought about it,
given
4Al + 302 --> 2Al2O3
and you have 0.185 M of Al,
and every 4 moles of Al goes to 2 moles of Al2O3, so (ignoring significant figures)
2 / 4 * 0.185 = 0.0925 moles of Al2O3
0.0925 * (26.98 * 2 + 3 * 16) = 9.43 g of Al2O3
My way of thinking is to convert everything to moles, and then work from there, do you see? Magpies answer is obviously that of an engineer: it's more
elegant! But I've always thought and worked from units.