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!
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.
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
To determine the volume of water that can be cleared of suspended solids by one beverage can, you need to calculate the mass of alum in one can and then use this information to find the concentration of alum in the water.
Based on the given data:
Mass of alum (KAl(SO4)2)*12H2O): 16.44 g
% yield = 95.41%
First, calculate the actual yield of alum:
Actual yield = % yield * Theoretical yield
Theoretical yield = Mass of Al = 0.98 g
Actual yield = 95.41% * 0.98 g = 0.9342818 g
Next, calculate the moles of alum:
Molar mass of alum = molar mass of KAl(SO4)2 * 12 H2O = (39.1 g/mol + 26.98 g/mol + 4 * 32.06 g/mol + 12 * 1.01 g/mol + 2 * 16 g/mol) * 12 = 474.392 g/mol
Moles of alum = Mass of alum / Molar mass of alum
Moles of alum = 0.9342818 g / 474.392 g/mol = 0.001968 mol
Now, assuming that all the alum dissolves completely in the water, the concentration of alum in the water can be calculated.
Concentration = Moles of solute / Volume of solution
Let's denote the volume of water as V.
Concentration = 0.001968 mol / V
Now, the concentration of suspended solids in the water after adding one beverage can is given as the mass of alum in one can divided by the volume of water cleared.
Mass of alum in one can = 16.44 g
Concentration of suspended solids = Mass of alum in one can / Volume of water cleared
Since we want to find the range of the volume of water cleared, we can rearrange the equation:
Volume of water cleared = Mass of alum in one can / Concentration of suspended solids
We already know the mass of alum in one can (16.44 g). Now, substitute the values:
Volume of water cleared = 16.44 g / (0.001968 mol / V)
After substituting, simplify the equation and calculate the range of the volume of water cleared using the given values of the mass of alum and the concentration of alum found earlier.