If the percent yield was 81.5%, how many grams of aluminum were used to prepare 20.0 g of
potassium alum?
The skeleton equation is
Al ==> KAl(SO4)2.12H2O
20g final product. 81.5% yield means
20/0.815 = 24.5 g KAl(SO4)2.12H2O formed.
mol = 24.5/molar mass KAl(SO4)2.12H2O
That many mol Al were required.
g Al = mols x atomic mass Al.
Well, I don't actually know how many grams of aluminum were used, but if we assume that the percent yield is related to how effective my jokes are, then I would say at least 81.5 grams of aluminum were used. Can't have a high yield without some serious material!
To determine the grams of aluminum used to prepare potassium alum, we need to use the concept of percent yield. Percent yield is a measure of how much of a desired product is obtained in a chemical reaction compared to the theoretical yield, which is the maximum possible yield.
The percent yield formula is:
Percent Yield = (Actual Yield / Theoretical Yield) x 100%
In this case, the percent yield is given as 81.5%. To find the grams of aluminum used, we need to calculate the theoretical yield of potassium alum. The theoretical yield is the amount of potassium alum that would be obtained if the reaction went to completion without any losses.
Since the molecular formula of potassium alum is KAl(SO4)2 · 12H2O, we know that the molar mass of KAl(SO4)2 · 12H2O is 474.388 g/mol.
To calculate the theoretical yield, we first need to determine the number of moles of potassium alum formed using the given mass of 20.0 g:
Number of Moles = Mass / Molar Mass = 20.0 g / 474.388 g/mol
Next, we need to determine the molar ratio of aluminum to potassium alum from the balanced chemical equation of the reaction used to prepare potassium alum. Let's assume the balanced equation is:
2Al + K2SO4 + 2H2O → 2KAl(SO4)2 + 3H2
From this equation, we can see that 2 moles of aluminum (2Al) are required to produce 1 mole of potassium alum (KAl(SO4)2).
Now, using the number of moles of potassium alum and the molar ratio of aluminum to potassium alum, we can calculate the number of moles of aluminum used:
Number of Moles of Aluminum = Number of Moles of Potassium Alum × (2 moles Al / 1 mole KAl(SO4)2)
Finally, to find the grams of aluminum used, we can multiply the number of moles of aluminum by the molar mass of aluminum (26.98 g/mol):
Grams of Aluminum = Number of Moles of Aluminum × Molar Mass of Aluminum
By following these steps, you can calculate the grams of aluminum used to prepare 20.0 g of potassium alum.