Calculate the mass of SbI₃ formed when 1.20 g of Sb react with 2.40 g of I₂, according to the equation,

2Sb_(s) + 3I_2(s) ---> 2SbI_3(s)

Solution:

To begin this problem we need to build the table.

In this table I decided to start with Sb.

(mol reactant Sb)o	(mol reactant I2)required	(mol reactant I2)o	Conclusion

To complete the table we must calculate the mols of Sb and I₂ initially,

Entering these values into the table we have,

(mol reactant Sb)o	(mol reactant I2)required	(mol reactant I2)o	Conclusion
9.84 x 10-3		9.45 x 10-3

Now we need to calculate the mols of I₂ required to completely react with 9.84 x 10^-3 mols Sb.

So the table now looks like;

(mol reactant Sb)o	(mol reactant I2)required	(mol reactant I2)o	Conclusion
9.84 x 10-3	1.48 x 10-2	9.45 x 10-3

The next step is determine the limiting reagent.

Looking at the the mols of I₂ required and comparing to the mols of I₂ initially we can determine whether I₂ is the limiting reagent or in excess.

(mol reactant Sb)o	(mol reactant I2)required	(mol reactant I2)o	Conclusion
9.84 x 10-3	1.48 x 10-2	9.45 x 10-3

In this case since the mols of I₂ required is greater than the mols of I₂ initially, we do not have enough I₂ to react with all the Sb, therefore I₂ is the limiting reagent.

(mol reactant Sb)o	(mol reactant I2)required	(mol reactant I2)o	Conclusion
9.84 x 10-3	1.48 x 10-2	9.45 x 10-3	I2 is limiting Sb is excess

To calculate the mass of SbI₃ formed we use the limiting reagent.