# A piece of an aluminum can was cut into small pieces. Then 0.422 g of the chips from the can was used to

prepare potassium alum according to the procedure we used. Calculate the theoretical yield (in g) of potassium
alum that could be obtained in the reaction. (Alum MW=474.41 g/mol).

## Potassium alum is KAl(SO4)2.12H2O

0.422g Al x (molar mass KAl(SO4)2.12H2O/atomic mass Al) = ? theoretical yield.

## To calculate the theoretical yield of potassium alum, we first need to determine the amount of aluminum present in the 0.422 g of aluminum chips.

1. Calculate the molar mass of aluminum (Al):
- Aluminum (Al) MW = 26.98 g/mol

2. Calculate the number of moles of aluminum:
- Number of moles = mass / molar mass
- Number of moles of Al = 0.422 g / 26.98 g/mol

Now, we can use stoichiometry to find the molar ratio between aluminum (Al) and potassium alum (KAl(SO4)2·12H2O). The balanced equation for the reaction is:

2Al + K2SO4·Al2(SO4)3·24H2O --> 2K[Al(SO4)2]·12H2O

From the balanced equation, we can see that 2 moles of aluminum (Al) react to produce 1 mole of potassium alum (KAl(SO4)2·12H2O).

3. Calculate the number of moles of potassium alum:
- Number of moles of KAl(SO4)2·12H2O = (Number of moles of Al) / 2

4. Calculate the theoretical yield of potassium alum in grams:
- Theoretical yield = (Number of moles of KAl(SO4)2·12H2O) × (Molar mass of KAl(SO4)2·12H2O)

Let's calculate the values:

1. Number of moles of Al = 0.422 g / 26.98 g/mol = 0.0157 mol

2. Number of moles of KAl(SO4)2·12H2O = (0.0157 mol Al) / 2 = 0.00786 mol

3. Theoretical yield = (0.00786 mol) × (474.41 g/mol) = 3.729 g

Therefore, the theoretical yield of potassium alum that could be obtained in the reaction is 3.729 grams.