**Calculating the Molecular Weight**

To solve this problem, we first need to calculate the molecular weight of the silver salt. The molecular weight is the sum of the atomic weights of all the atoms in the molecule.

The atomic weight of silver (Ag) is given as 108 g/mol. Since we have 0.5 mole of the silver salt, we can calculate the weight of the silver salt using the equation:

Weight of silver salt = number of moles * molecular weight

Weight of silver salt = 0.5 mol * 108 g/mol = 54 g

**Determining the Percent Yield**

The weight of the residue obtained is given as 216 g. To determine the percent yield, we need to compare the actual yield (weight of the residue) to the theoretical yield (weight of silver salt).

The percent yield can be calculated using the equation:

Percent yield = (actual yield / theoretical yield) * 100

The theoretical yield is the weight of the silver salt, which we calculated as 54 g.

Percent yield = (216 g / 54 g) * 100 = 400%

**Analyzing the Result**

A percent yield of 400% is not possible in a chemical reaction. The percent yield represents the efficiency of a reaction and is a measure of how much product is obtained compared to the theoretical yield.

A percent yield greater than 100% indicates that there is an error in the calculation or an additional source of the product. In this case, the weight of the residue obtained is higher than the weight of the silver salt used.

**Possible Explanations**

There are several possible explanations for a percent yield greater than 100%:

1. Contamination: The residue obtained may contain impurities, leading to a higher weight than expected.

2. Additional Reaction: It is possible that another reaction occurred during the experiment, producing additional product.

3. Experimental Error: There may have been errors in the measurements or calculations, leading to an inaccurate percent yield.

**Conclusion**

Based on the given information, a percent yield of 400% is not possible. It is likely that there were errors in the experiment or additional sources of product. Further investigation and analysis are needed to determine the exact cause of the high percent yield.