Calculate the percent yield of alum if you obtained 75.0 g of potassium alum from 5.00 g Al.
(75/87.7848)*100=85.4362%
Skeleton equation
Al ==> KAl(SO4)2.12H2O
mols Al = 5g/atomic mass Al
mols KAl(SO4)2.12H2O moles Al.
g KAl(SO4)2.12H2O = moles x molar mass. This is the theoretical yield.
%yield = (75/theoretical yield)*100= ?
Just use the formula from "DaDood FrumCheers" Theorem which is N=Pv(Pv2)/T.
To calculate the percent yield of alum, you need to know both the actual yield (the amount of alum obtained) and the theoretical yield (the amount of alum that should have been obtained based on stoichiometry).
To find the theoretical yield, you need to balance the chemical equation for the reaction between aluminum and potassium sulfate to form alum. The balanced equation is:
2 Al + K2SO4 + 4 H2O → KAl(SO4)2·12H2O + H2
From the balanced equation, you can see that 2 moles of aluminum react with 1 mole of potassium sulfate to produce 1 mole of potassium alum.
First, we need to convert the given mass of aluminum (5.00 g) to moles. The molar mass of aluminum (Al) is 26.98 g/mol. Using the formula:
moles = mass / molar mass
moles of Al = 5.00 g / 26.98 g/mol = 0.185 mol
Since the stoichiometry of aluminum to potassium alum is 2:1, the number of moles of alum formed will be half of the number of moles of aluminum.
moles of KAl(SO4)2·12H2O = 0.185 mol / 2 = 0.0925 mol
Now, we can calculate the theoretical yield of alum. The molar mass of KAl(SO4)2·12H2O is 474.39 g/mol.
theoretical yield = moles of KAl(SO4)2·12H2O × molar mass of KAl(SO4)2·12H2O
theoretical yield = 0.0925 mol × 474.39 g/mol = 43.84 g
Now that we have the actual yield (75.0 g) and the theoretical yield (43.84 g), we can calculate the percent yield using the formula:
percent yield = (actual yield / theoretical yield) × 100
percent yield = (75.0 g / 43.84 g) × 100 = 171.2%
Therefore, the percent yield of alum is 171.2%.