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How much alum can be prepared from a given mass of aluminum?
- If you start with 1.13g of aluminum, what mass of alum can theoretically be prepared? These are the formulas given to me...I'm not sure which one to use. 2Al + 2KOH + 6H2O 2 KAl(OH)4 + 3 H2
KAl(OH)4 + 2H2SO4 4 H2O + K+ + Al3+ + 2SO42- This is the formula for alum - KAl(SO4)2*12H2O
The equations show you how to prepare alum from aluminum. But to find the maximum amount
of alum that can be prepared from an amount of aluminum, all you really need is the
mole-to-mole ratio that relates alum and aluminum. You can get this from the formula
for alum: every mole of alum contains one mole of aluminum.
Follow this strategy when attacking problems of this type:
- What am I trying to find? You want grams of alum.
- What information is given? You have 1.13 g of Al, and you know the formula
- Connect the information given with what you're trying to find.
The formula of alum tells you that one mole of alum contains one mole of Al.
It also tells you how many grams of alum you have in one mole of alum (add up the atomic
weights). You know that Al's atomic weight can be used to connect grams of Al with
moles of Al. Linking these relationships, you'll do the following operations:
|1.13 g Al
||atomic weight of Al
||Al to alum
||molecular weight of alum
- Do the math.
- Is the answer reasonable? The mass of the alum should be larger than the amount of aluminum, right?
Author: Fred Senese firstname.lastname@example.org