# I'm working on a first year chemistry lab where we synthesized alum from alunimini foil and I'm doing the report and one of the questions is as follows:

Based on your findings, what volume of water can be cleared of suspended solids by one beverage can (give a range in L)? (Assume the mass of the beverage can to be 13.33 g)

Mass of Al: 0.98 g
Mass of alum (KAl(SO4)2)*12H2O): 16.44 g
% yield = 95.41%

Any help as to where I can start would be very much appreciated!

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1. According to your percentage yield, the beverage can can produce 12.72g of alum

> 13.33 * (95.41%) = 12.72g

If you refer to page 49 of your laboratory book, under the "the laboratory project", you will see written:

"25g to 300g of alum per 1000L of water is required to remove suspended solids."

To find the Volume of water per 12.72g of alum that is required to remove suspended solids, divide each mass by your mass:

25/12.72 = 1.97
300/12.72 = 23.58

and then divide 1000L of water by each number:

1000/1.97 = 508.8L
1000/23.58 = 42.4L

Hence, 12.72g of alum can clear 42.2 - 508.8 L of water from suspended solids.

Hope this helped.

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2. I'm pretty sure the 13.33g is mass of aluminum not alum. You found that 0.92g of aluminum gives 16.44g of alum so how much g of alum do u get from 13.33g of aluminum

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3. Hi I am 10. RP pls.

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4. Hi I am 10. RP pls. Kimochi

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